How many ways can the 7 letters mn o p q r s be arranged so that P and Q occupy continuous positions?

The word ARRANGEMENT has $11$ letters, not all of them distinct. Imagine that they are written on little Scrabble squares. And suppose we have $11$ consecutive slots into which to put these squares.

There are $\dbinom{11}{2}$ ways to choose the slots where the two A's will go. For each of these ways, there are $\dbinom{9}{2}$ ways to decide where the two R's will go. For every decision about the A's and R's, there are $\dbinom{7}{2}$ ways to decide where the N's will go. Similarly, there are now $\dbinom{5}{2}$ ways to decide where the E's will go. That leaves $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a total of $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$

Factorial Notation -> ! ; n! = n(n – 1) (n – 2) ... 3.2.1 = Product of n consecutive integers starting from 1.

0! = 1

Factorials of only Natural numbers are defined. n! is defined only for n ≥ 0 n! is not defined for n < 0

nCr = 1 when n = r.

Combinations (represented by nCr) can be defined as the number of ways in which r things at a time can be SELECTED from amongst n things available for selection. The key word here is SELECTION.

Understand here that the order in which the r things are selected has no importance in the counting of combinations.

nCr = Number of combinations (selections) of n things taken r at a time. nCr = n! /[r! (n–r)!] ; where n ≥ r (n is greater than or equal to r).

Some typical situations where selection/combination is used:

Selection of people for a team, a party, a job, an office etc. (e.g. Selection of a cricket team of 11 from 16 members)

Selection of a set of objects (like letters, hats, points pants, shirts, etc) from amongst another set available for selection.

In other words any selection in which the order of selection holds no importance is counted by using combinations.

Permutations (represented by nPr) can be defined as the number of ways in which r things at a time can be SELECTED & ARRANGED at a time from amongst n things.

The key word here is ARRANGEMENT. Hence please understand here that the order in which the r things are arranged has critical importance in the counting of permutations.

In other words permutations can also be referred to as an ORDERED SELECTION. nPr = number of permutations (arrangements) of n things taken r at a time.

nPr = n!/ (n – r)!; n ≥ r

Some typical situations where ordered selection/ permutations are used:

Making words and numbers from a set of available letters and digits respectively

Filling posts with people

Selection of batting order of a cricket team of 11 from 16 members

Putting distinct objects/people in distinct places, e.g. making people sit, putting letters in envelopes, finishing order in horse race, etc.)

Selection : Suppose we have three men A, B and C out of which 2 men have to be selected to two posts. This can be done in the following ways: AB, AC or BC (These three represent the basic selections of 2 people out of three which are possible. Physically they can be counted as 3 distinct selections. This value can also be got by using 3C2. Note here that we are counting AB and BA as one single selection. So also AC and CA and BC and CB are considered to be the same instances of selection since the order of selection is not important.

Arrangement : Suppose we have three men A, B and C out of which 2 men have to be selected to the post of captain and vice captain of a team. In this case we have to take AB and BA as two different instances since the order of the arrangement makes a difference in who is the captain and who is the vice captain. Similarly, we have BC and CB and AC and CA as 4 more instances. Thus in all there could be 6 arrangements of 2 things out of three. This is given by 3P2 = 6.

When we look at the formulae for Permutations and Combinations and compare the two we see that, nPr = r! × nCr = nCr × rPr

This in words can be said as: The permutation or arrangement of r things out of n is nothing but the selection of r things out of n followed by the arrangement of the r selected things amongst themselves

MNP Rule : If there are three things to do and there are M ways of doing the first thing, N ways of doing the second thing and P ways of doing the third thing then there will be M × N × P ways of doing all the three things together.The works are mutually inclusive.

This is used to for situations like: The numbers 1, 2, 3, 4 and 5 are to be used for forming 3 digit numbers without repetition. In how many ways can this be done?

Using the MNP rule you can visualise this as: There are three things to doÆ The first digit can be selected in 5 distinct ways, the second can be selected in 4 ways and the third can be selected in 3 different ways. Hence, the total number of 3 digit numbers that can be formed are 5 × 4 × 3 = 60

When the pieces of work are mutually exclusive, there are M+N+P ways of doing the complete work.

Number of permutations (or arrangements) of n different things taken all at a time = n!

Number of permutations of n things out of which P1 are alike and are of one type, P2 are alike and are of a second type and P3 are alike and are of a third type and the rest are all different = $ \frac{n!}{P1! * P2! * P3!} $

Illustration: The number of words formed with the letters of the word Allahabad. Solution: Total number of Letters = 9 of which A occurs four times, L occurs twice and the rest are all different. Total number of words formed = 9! / (4! 2! 1! )

Number of permutations of n different things taken r at a time when repetition is allowed = n × n × n × ... (r times) = nr .

Illustration: In how many ways can 4 rings be worn in the index, ring finger and middle finger if there is no restriction of the number of rings to be worn on any finger? Solution: Each of the 4 rings could be worn in 3 ways either on the index, ring or middle finger. So, four rings could be worn in 3 × 3 × 3 × 3 = 34 ways.

Number of selections of r things out of n identical things = 1 Illustration: In how many ways 5 marbles can be chosen out of 100 identical marbles? Solution: Since, all the 100 marbles are identical Hence, Number of ways to select 5 marbles = 1

Total number of selections of zero or more things out of k identical things = k + 1. This includes the case when zero articles are selected.

Total number of selections of zero or more things out of n different things $= nC_0 + nC_1 + ... + nC_n = 2^n$

Corollary: The number of selections of 1 or more things out of n different things = $= nC_1 + ... + nC_n = 2^n - 1$

Number of ways of distributing n identical things among r persons when each person may get any number of things = n + r – 1Cr–1

Imagine a situation where 27 marbles have to be distributed amongst 4 people such that each one of them can get any number of marbles (including zero marbles). Then for this situation we have, n = 27(no. of identical objects), r = 4 (no. of people) and the answer of the number of ways this can be achieved is given by: n + r – 1Cr–1 = 30C3

Corollary: No. of ways of dividing n non distinct things to r distinct groups are: n–1Cr–1 so For non-empty groups only Also, the number of ways in which n distinct things can be distributed to r different persons: = rn

Number of ways of dividing m + n different things in two groups containing m and n things respectively = m + nCn × mCm = = (m + n)! /m! n! Or, m+nCm × nCn = (m + n)! / n! m!

Number of ways of dividing 2n different things in two groups containing n things = 2n! / n! n! 2!

nCr + nCr – 1 = n + 1Cr

nCx = nCy so x = y or x + y = n

nCr = nCn – r

r . nCr = n . n – 1Cr – 1

nCr / (r + 1) = n + 1Cr + 1/( n + 1)

For nCr to be greatest,

if n is even, r = n/2

if n is odd, r = (n + 1)/2 or (n – 1)/2

Number of selections of r things out of n different things

When k particular things are always included = n – kCr – k

When k particular things are excluded = n – kCr

When all the k particular things are not together in any selection = nCr – n – kCr – k

No. of ways of doing a work with given restriction = total no. of ways of doing it — no. of ways of doing the same work with opposite restriction.

The total number of ways in which 0 to n things can be selected out of n things such that p are of one type, q are of another type and the balance r of different types is given by: (p + 1)(q + 1)(2r– 1).

Total number of ways of taking some or all out of p + q + r things such that p are of one type and q are of another type and r of a third type = (p + 1)(q + 1)(r + 1) – 1 [Only non-empty sets]

$ \frac{nCr}{nCr-1} = \frac{n-r+1}{r} $

Number of selections of k consecutive things out of n things in a row = n – k + 1

Consider two situations: There are three A, B and C. In the first case, they are arranged linearly and in the other, around a circular table

For the linear arrangement, each arrangement is a totally new way. For circular arrangements, three linear arrangements are represented by one and the same circular arrangement.

So, for six linear arrangements, there correspond only 2 circular arrangements. This happens because there is no concept of a starting point on a circular arrangement. (i.e., the starting point is not defined.) Generalising the whole process, for n!, there corresponds to be (1/n) n! ways.

Number of ways of arranging n people on a circular track (circular arrangement) = (n – 1)!

When clockwise and anti-clockwise observation are not different then number of circular arrangements of n different things = (n – 1)! /2 e.g. the case of a necklace with different beads, the same arrangement when looked at from the opposite side becomes anti-clockwise.

Number of selections of k consecutive things out of n things in a circle = n when k < n = 1 when k = n

Number of terms in $ (a_1 + a_2 + ... + a_n)^m$ is m + n – 1Cn–1

Q. Find the number of terms in (a + b + c)2

n = 3, m = 2

m + n – 1Cn–1 = 4C2 = 6

Corollary: Number of terms in $ [1 + x + x^2 v+ ... + x^n]^m = mn+1 $

Q. Find the number of zeroes at the end of 1000!

Number of zeroes ending the number represented by n! = [n/5] + [n /52] + [n/53] + ... + [n/5x]

[] Shows greatest integer function where 5x ≤ n

[1000/5] + [1000 /52] + [1000/53] + ... + [1000/54]

200 + 40 + 8 + 1 = 249

Q. Find how many exponents of 3 will be there in 24!.

Find how many exponents of 3 is represented by n! = [n/3] + [n /32] + [n/33] + ... + [n/3x]

[] Shows greatest integer function where 3x ≤ n

[24/3] + [24/32] = 8 + 2 = 10

Example : The number of squares and rectangles in the figure are given by:

Number of squares in a square of n × n side = $1^2 + 2^2 + 3^2 + ... + n^2$

Number of rectangles in a square of n × n side = $1^3 + 2^3 + 3^3 + ... + n^3$ . (This includes the number of squares.)

A rectangle having m rows and n columns: The number of squares is given by: m.n + (m – 1)(n – 1) + (m – 2)(n – 2) + ... until any of (m – x) or (n – x) comes to 1

The number of rectangles is given by: (1 + 2 + ... + m)(1 + 2 + ... + n)

Q. The number of squares and rectangles in the following figure are given by:

Number of squares = $1^2 + 2^2 + 3^2 = 14 $

Number of rectangles = $ 1^3 + 2^3 + 3^3 = 36 $

Find the number of permutations of 6 things taken 4 at a time.

Explanation :The answer will be given by 6P4

How many 3-digit numbers can be formed out of the digits 1, 2, 3, 4 and 5?

Explanation :Forming numbers requires an ordered selection. Hence, the answer will be 5P3 .

In how many ways can the 7 letters M, N, O, P, Q, R, S be arranged so that P and Q occupy continuous positions?

Explanation :n For arranging the 7 letters keeping P and Q always together we have to view P and Q as one letter. Let this be denoted by PQ. Then, we have to arrange the letters M, N, O, PQ, R and S in a linear arrangement. Here, it is like arranging 6 letters in 6 places (since 2 letters are counted as one). This can be done in 6! ways. However, the solution is not complete at this point of time since in the count of 6! the internal arrangement between P and Q is neglected. This can be done in 2! ways. Hence, the required answer is 6! × 2!.

What would happen if the letters P, Q and R are to be together? in above question?

What if P and Q are never together? in above question ?

Explanation :(Answer will be given by the formula: Total number of ways – Number of ways they are always together)

Of the different words that can be formed from the letters of the words BEGINS how many begin with B and end with S?

Explanation :B & S are fixed at the start and the end positions. Hence, we have to arrange E, G, I and N amongst themselves. This can be done in 4! ways.

What will be the number of words that can be formed with the letters of the word BEGINS which have B and S at the extreme positions?

In how many ways can the letters of the word VALEDICTORY be arranged, so that all the vowels are adjacent to each other?

Explanation :There are 4 vowels and 7 consonants in Valedictory. If these vowels have to be kept together, we have to consider AEIO as one letter. Then the problem transforms itself into arranging 8 letters amongst themselves (8! ways). Besides, we have to look at the internal arrangement of the 4 vowels amongst themselves. (4! ways) Hence Answer = 8! × 4!

If there are two kinds of hats, red and blue and at least 5 of each kind, in how many ways can the hats be put in each of 5 different boxes?

Explanation :The significance of at least 5 hats of each kind is that while putting a hat in each box, we have the option of putting either a red or a blue hat. (If this was not given, there would have been an uncertainty in the number of possibilities of putting a hat in a box.) Thus in this question for every task of putting a hat in a box we have the possibility of either putting a red hat or a blue hat. The solution can then be looked at as: there are 5 tasks each of which can be done in 2 ways. Through the MNP rule we have the total number of ways = 25 (Answer).

In how many ways can 4 Indians and 4 Nepalese people be seated around a round table so that no two Indians are in adjacent positions?

Explanation :If we first put 4 Indians around the round table, we can do this in 3! ways. Once the 4 Indians are placed around the round table, we have to place the four Nepalese around the same round table. Now, since the Indians are already placed we can do this in 4! ways (as the starting point is defined when we put the Indians. Try to visualize this around a circle for placing 2 Indians and 2 Nepalese.) Hence, Answer = 3! × 4!

How many numbers greater than a million can be formed from the digits 1, 2, 3, 0, 4, 2, 3?

Explanation :In order to form a number greater than a million we should have a 7 digit number. Since we have only seven digits with us we cannot take 0 in the starting position. View this as 7 positions to fill: _ _ _ _ _ _ _ To solve this question we first assume that the digits are all different. Then the first position can be filled in 6 ways (0 cannot be taken), the second in 6 ways (one of the 6 digits available for the first position was selected. Hence, we have 5 of those 6 digits available. Besides, we also have the zero as an additional digit), the third in 5 ways (6 available for the 2nd position – 1 taken for the second position.) and so on. Mathematically this can be written as: 6 × 6 × 5 × 4 × 3 × 2 × 1 = 6 × 6! This would have been the answer had all the digits been distinct. But in this particular example we have two 2’s and two 3’s which are identical to each other. This complication is resolved as follows to get the answer: (6 * 6)/(2! * 2!)

If there are 11 players to be selected from a team of 16, in how many ways can this be done?

In how many ways can 18 identical white and 16 identical black balls be arranged in a row so that no two black balls are together?

Explanation :When 18 identical white balls are put in a straight line, there will be 19 spaces created. Thus 16 black balls will have 19 places to fill in. This will give an answer of: 19C16 . (Since, the balls are identical the arrangement is not important.)

A mother with 7 children takes three at a time to a cinema. She goes with every group of three that she can form. How many times can she go to the cinema with distinct groups of three children?

Explanation : She will be able to do this as many times as she can form a set of three distinct children from amongst the seven children. This essentially means that the answer is the number of selections of 3 people out of 7 that can be done. Hence, Answer = 7C3

For the above question, how many times will an individual child go to the cinema with her before a group is repeated?

Explanation :This can be viewed as: The child for whom we are trying to calculate the number of ways is already selected. Then, we have to select 2 more children from amongst the remaining 6 to complete the group. This can be done in 6C2 ways

How many different sums can be formed with the following coins: 5 rupee, 1 rupee, 50 paisa, 25 paisa, 10 paisa and 1 paisa?

Explanation :A distinct sum will be formed by selecting either 1 or 2 or 3 or 4 or 5 or all 6 coins. But from the formula we have the answer to this as : 26 – 1. [Task for the student: How many different sums can be formed with the following coins: 5 rupee, 1 rupee, 50 paisa, 25 paisa, 10 paisa, 3 paisa, 2 paisa and 1 paisa? Hint: You will have to subtract some values for double counted sums.]

A train is going from Mumbai to Pune and makes 5 stops on the way. 3 persons enter the train during the journey with 3 different tickets. How many different sets of tickets may they have had?

Explanation :Since the 3 persons are entering during the journey they could have entered at the: 1st station (from where they could have bought tickets for the 2nd, 3rd, 4th or 5th stations or for Pune total of 5 tickets.) 2nd station (from where they could have bought tickets for the 3rd, 4th or 5th stations or for Pune Æ total of 4 tickets.) 3rd station (from where they could have bought tickets for the 4th or 5th stations or for Pune Æ total of 3 tickets.) 4th station (from where they could have bought tickets for the 5th station or for Pune Æ total of 2 tickets.) 5th station (from where they could have bought a ticket for Pune Æ total of 1 ticket.) Thus, we can see that there are a total of 5 + 4 + 3 + 2 + 1 = 15 tickets available out of which 3 tickets were selected. This can be done in 15C3 ways (Answer).

Find the number of diagonals and triangles formed in a decagon.

Explanation :A decagon has 10 vertices. A line is formed by selecting any two of the ten vertices. This can be done in 10C2 ways. However, these 10C2 lines also count the sides of the decagon. Thus, the number of diagonals in a decagon is given by: 10C2 – 10 (Answer) Triangles are formed by selecting any three of the ten vertices of the decagon. This can be done in 10C3 ways (Answer).

Out of 18 points in a plane, no three are in a straight line except 5 which are collinear. How many straight lines can be formed?

Explanation :If all 18 points were non-collinear then the answer would have been 18C2 . However, in this case 18C2 has double counting since the 5 collinear points are also amongst the 18. These would have been counted as 5C2 whereas they should have been counted as 1. Thus, to remove the double counting and get the correct answer we need to adjust by reducing the count by (5C2 – 1). Hence, Answer =18C2 – (5C2 – 1) = 18C2 – 5C2 + 1

For the above situation, how many triangles can be formed?

Explanation :The triangles will be given by 18C3 – 5C3.

A question paper had ten questions. Each question could only be answered as True (T) or False (F). Each candidate answered all the questions. Yet, no two candidates wrote the answers in an identical sequence. How many different sequences of answers are possible?

Explanation :210 = 1024 unique sequences are possible. Option (d) is correct

When ten persons shake hands with one another, in how many ways is it possible?

Explanation :For n people there are always nC2 shake hands. Thus, for 10 people shaking hands with each other the number of ways would be 10C2 = 45.

In how many ways can four children be made to stand in a line such that two of them, A and B are always together?

Explanation :If the children are A, B, C, D we have to consider A & B as one child. This, would give us 3! ways of arranging AB, C and D. However, for every arrangement with AB, there would be a parallel arrangement with BA. Thus, the correct answer would be 3! × 2! = 12 ways. Option (b) is correct

Each person’s performance compared with all other persons is to be done to rank them subjectively. How many comparisons are needed to total, if there are 11 persons?

Explanation :There would be 11C2 combinations of 2 people taken 2 at a time for comparison. 11C2 = 55

A person X has four notes of Rupee 1, 2, 5 and 10 denomination. The number of different sums of money she can form from them is

Explanation :24 – 1 = 15 sums of money can be formed. Option (b) is correct

A person has 4 coins each of different denomination. What is the number of different sums of money the person can form (using one or more coins at a time)?

Explanation : 24 – 1 = 15. Hence, option (b) is correct.

How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9, such that the digits are in ascending order?

Explanation :Numbers starting with 12 – 7 numbers Numbers starting with 13 – 6 numbers; 14 – 5, 15 – 4, 16 – 3, 17 – 2, 18 – 1. Thus total number of numbers starting from 1 is given by the sum of 1 to 7 = 28. Number of numbers starting from 2- would be given by the sum of 1 to 6 = 21 Number of numbers starting from 3- sum of 1 to 5 = 15 Number of numbers starting from 4 – sum of 1 to 4 = 10 Number of numbers starting from 5 – sum of 1 to 3 = 6 Number of numbers starting from 6 = 1 + 2 = 3 Number of numbers starting from 7 = 1 Thus a total of: 28 + 21 + 15 + 10 + 6 + 3 + 1 = 84 such numbers. Option (d) is correct

In a carrom board game competition, m boys n girls (m > n > 1) of a school participate in which every student has to play exactly one game with every other student. Out of the total games played, it was found that in 221 games one player was a boy and the other player was a girl. Consider the following statements: I. The total number of students that participated in the competition is 30. II. The number of games in which both players were girls is 78.

Which of the statements given above is/are correct?

Explanation :The given condition can get achieved if we were to use 17 boys and 13 girls. In such a case both statement I and II are correct. Hence, option (c) is correct.

In how many different ways can all of 5 identical balls be placed in the cells shown above such that each row contains at least 1 ball?

Explanation : The placement of balls can be 3, 1, 1 and 2, 2, 1. For 3, 1, 1- If we place 3 balls in the top row, there would be 3C1 ways of choosing a place for the ball in the second row and 3C1 ways of choosing a place for the ball in the third row. Thus, 3C1 × 3C1 = 9 ways. Similarly there would be 9 ways each if we were to place 3 balls in the second row and 3 balls in the third row. Thus, with the 3, 1, 1 distribution of 5 balls we would get 9 + 9 + 9 = 27 ways of placing the balls. We now need to look at the 2, 2, 1 arrangement of balls. If we place 1 ball in the first row, we would need to place 2 balls each in the second and the third rows. In such a case, the number of ways of arranging the balls would be 3C1 × 3C2 × 3C2 = 27 ways. (choosing 1 place out of 3 in the first row, 2 places out of 3 in the second row and 2 places out of 3 in the third row). Similarly if we were to place 1 ball in the second row and 2 balls each in the first and third rows we would get 27 ways of placing the balls and another 27 ways of placing the balls if we place 1 ball in the third row and 2 balls each in the other two rows. Thus with a 2, 2, 1 distribution of the 5 balls we would get 27 + 27 + 27 = 81 ways of placing the balls. Hence, total number of ways = Number of ways of placing the balls with a 3,1,1 distribution of balls + number of ways of placing the balls with a 2, 2, 1 distribution of balls = 27 + 81 = 108. Hence, option (d) is correct.

There are 6 different letter and 6 correspondingly addressed envelopes. If the letters are randomly put in the envelopes, what is the probability that exactly 5 letters go into the correctly addressed envelopes?

Explanation :If 5 letters go into the correct envelopes the sixth would automatically go into it’s correct envelope. Thus, there is no possibility when exactly 5 letters are correct and 1 is wrong. Hence, option (a) is correct.

There are two identical red, two identical black and two identical white balls. In how many different ways can the balls be placed in the cells (each cell to contain one ball) shown above such that balls of the same colour do not occupy any two consecutive cells?

Explanation :In the first cell, we have 3 options of placing a ball. Suppose we were to place a red ball in the first cell- then the second cell can only be filled with either black or white – so 2 ways. Subsequently there would be 2 ways each of filling each of the cells (because we cannot put the colour we have already used in the previous cell). Thus, the required number of ways would be 3 × 2 × 2 × 2 = 24 ways. Hence, option (c) is correct.

How many different triangles are there in the figure shown above?

Explanation :Look for the smallest triangles first—there are 12 of them. Then, look for the triangles which are equal to half the rectangle—there are 12 of them. Besides, there are 4 bigger triangles (spanning across 2 rectangles). Thus a total of 28 triangles can be seen in the figure. Hence, option (a) is correct

A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included a particular student?

Explanation :If the students are A, B, C, D, E and F- we can have 6C3 groups in all. However, if we have to count groups in which a particular student (say A) is always selected- we would get 5C2 = 10 ways of doing it. Hence, option (c) is correct

Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2?

Explanation :All 3 dice have twos – 1 case. Two dice have twos: This can principally occur in 3 ways which can be broken into: If the first two dice have 2- the third dice can have 1, 3, 4, 5 or 6 = 5 ways. Similarly, if the first and third dice have 2, the second dice can have 5 outcomes Æ 5 ways and if the second and third dice have a 2, there would be another 5 ways. Thus a total of 15 outcomes if 2 dice have a 2. With only 1 dice having a two- If the first dice has 2, the other two can have 5 × 5 = 25 outcomes. Similarly 25 outcomes if the second dice has 2 and 25 outcomes if the third dice has 2. A total of 75 outcomes. Thus, a total of 1 + 15 + 75 = 91 possible outcomes with at least. Hence, option (c) is correct.

All the six letters of the name SACHIN are arranged to form different words without repeating any letter in any one word. The words so formed are then arranged as in a dictionary. What will be the position of the word SACHIN in that sequence?

Explanation :All words staring with A, C, H, I and N would be before words starting with S. So we would have 5! Words (= 120 words) each starting with A, C, H, I and N. Thus, a total of 600 words would get completed before we start off with S. SACHIN would be the first word starting with S, because A, C, H, I, N in that order is the correct alphabetical sequence. Hence, Sachin would be the 601 st word. Hence, option (c) is correct.

Five balls of different colours are to be placed in three different boxes such that any box contains at least one 1 ball. What is the maximum number of different ways in which this can be done?

Explanation :The arrangements can be [3 & 1 & 1 or 1 & 3 & 1 or 1 & 1 & 3] or 2 & 2 & 1 or 2 & 1 & 2 or 1 & 2 and 2. Total number of ways = 3 × 5C3 × 2C1 × 1C1 + 3 × 5C2 × 3C2 × 1C1 = 60 + 90 = 150 ways Hence, option (c) is correct.

Amit has five friends: 3 girls and 2 boys. Amit’s wife also has 5 friends : 3 boys and 2 girls. In how many maximum number of different ways can they invite 2 boys and 2 girls such that two of them are Amit’s friends and two are his wife’s?

Explanation :n The selection can be done in the following ways: 2 boys from Amit’s friends and 2 girls from his wife’s friends OR 1 boy & 1 girl from Amit’s friends and 1 boy and 1 girl from his wife’s friends OR 2 girls from Amit’s friends and 2 boys from his wife’s friends. The number of ways would be: 2C2 × 2C2 + 3C1 × 2C1 × 3C1 × 2C1 + 3C2 × 3C2 = 1 + 36 + 9 = 46 ways.

In the given figure, what is the maximum number of different ways in which 8 identical balls can be placed in the small triangles 1, 2, 3 and 4 such that each triangle contains at least one ball?

Explanation :The ways of placing the balls would be 5, 1, 1, 1 (4!/3! = 4 ways); 4, 2, 1 & 1 (4!/2! = 12 ways); 3, 3, 1, 1 (4!/2! × 2! = 6 ways); 3, 2, 2, 1 (4!/2! = 12 ways) and 2, 2, 2, 2 (1 way). Total number of ways = 4 + 12 + 6 + 12 + 1 = 35 ways. Hence, option (b) is correct.

6 equidistant vertical lines are drawn on a board. 6 equidistant horizontal lines are also drawn on the board cutting the 6 vertical lines, and the distance between any two consecutive horizontal lines is equal to that between any two consecutive vertical lines. What is the maximum number of squares thus formed?

Explanation :The number of squares would be 12 + 22 + 32 + 42 + 52 + 62 = 91. Hence, option (c) is correct

Groups each containing 3 boys are to be formed out of 5 boys—A, B, C, D and E such that no group contains both C and D together. What is the maximum number of such different groups?

Explanation :All groups – groups with C and D together = 5C3 – 3C1 = 10 – 3 = 7

In how many maximum different ways can 3 identical balls be placed in the 12 squares (each ball to be placed in the exact centre of the squares and only one ball is to be placed in one square) shown in the figure given above such that they do not lie along the same straight line ?

Explanation :The thought process for this question would be: All arrangements (12C3) – Arrangements where all 3 balls are in the same row (3 × 4C3) – arrangements where all 3 balls are in the same straight line diagonally (4 arrangements) = 12C3 – 3 × 4C3 – 4 = 220 – 12 – 4 = 204 ways. Hence, option (c) is correct

How many numbers are there in all from 6000 to 6999 (Both 6000 and 6999 included) having at least one of their digits repeated?

Explanation :All numbers – numbers having no numbers repeated = 1000 – 9 × 8 × 7 = 1000 – 504 = 496 numbers. Hence, option (c) is correct

Each of two women and three men is to occupy one chair out of eight chairs, each of which is numbered from one to eight. First, women are to occupy any two chairs from\those numbered one to four; and then the three men would occupy any three chairs out of the remaining six chairs. What is the maximum number of different ways in which this can be done?

Explanation :4C2 × 2! × 6C3 × 3! = 6 × 2 × 20 × 6 = 1440. Hence, option (c) is correct

A box contains five set of balls while there are three balls in each set. Each set of balls has one colour which is different from every other set. What is the least number of balls that must be removed from the box in order to claim with certainty that a pair of balls of the same colour has been removed?

Explanation :Let C1, C2, C3, C4 and C5 be the 5 distinct colours which have no repetition. For being definitely sure that we have picked up 2 balls of the same colour we need to consider the worst case situation.
In the above distribution of balls each set has exactly 1 ball which is unique in it’s colour while the colours of the other two balls are shared at least once in one of the other sets. In such a case, the worst scenario would be if we pick up the first 10 balls and they all turn out to be of different colours. The 11 th ball has to be of a colour which has already been taken. Thus,if we were to pick out 11 balls we would be sure of having at least 2 balls of the same colour. Hence, option (d) is correct.

In a question paper, there are four multiple-choice questions. Each question has five choices with only one choice as the correct answer. What is the total number of ways in which a candidate will not get all the four answers correct?

Explanation : 54 would be the total number of ways in which the questions can be answered. Out of these there would be only 1 way of getting all 4 correct. Thus, there would be 624 ways of not getting all answers correct.

Each of 8 identical balls is to be placed in the squares shown in the figure given in a horizontal direction such that one horizontal row contains 6 balls and the other horizontal row contains 2 balls. In how many maximum different ways can this be done?

Explanation :The 6 balls must be on either of the middle rows. This can be done in 2 ways. Once, we put the 6 balls in their single horizontal row- it becomes evident that for placing the 2 remaining balls on a straight line there are 2 principal options: 1. Placing the two balls in one of the four rows with two squares. In this case the number of ways of placing the balls in any particular row would be 1 way (since once you were to choose one of the 4 rows, the balls would automatically get placed as there are only two squares in each row.) Thus the total number of ways would be 2 × 4 × 1 = 8 ways. 2. Placing the two balls in the other row with six squares. In this case the number of ways of placing the 2 balls in that row would be 6C2 . This would give us 2C1 × 1 × 6C2 = 30 ways. Total is 30 + 8 = 38 ways. Hence, option (a) is correct

In a tournament each of the participants was to play one match against each of the other participants. 3 players fell ill after each of them had played three matches and had to leave the tournament. What was the total number of participants at the beginning, if the total number of matches played was 75?

Explanation :The number of players at the start of the tournament cannot be 8, 10 or 12 because in each of these cases the total number of matches would be less than 75 (as 8C2,10C2 and12C2 are all less than 75.) This only leaves 15 participants in the tournament as the only possibility. Hence, option (d) is correct.

There are three parallel straight lines. Two points A and B are marked on the first line, points C and D are marked on the second line and points E and F are marked on the third line. Each of these six points can move to any position on its respective straight line. Consider the following statements: I. The maximum number of triangles that can be drawn by joining these points is 18. II. The minimum number of triangles that can be drawn by joining these points is zero.

Which of the statements given above is/are correct?

Explanation :The maximum triangles would be in case all these 6 points are non-collinear. In such a case the number of triangles is 6C3 = 20. Statement I is incorrect. Statement II is correct because if we take the position that A and B coincide on the first line, C & D coincide on the second line, E & F coincide on the third line and all these coincidences happen at 3 points which are on the same straight line- in such a case there would be 0 triangles formed. Hence, option (b) is correct

A mixed doubles tennis game is to be played between two teams (each team consists of one male and one female). There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

Explanation :First select the two men. This can be done in 4C2 ways. Let us say the men are A, B, C and D and their respective wives are a, b, c and d. If we select A and B as the two men then while selecting the women there would be two cases as seen below:
Total number of ways of doing this = 4C2 × 2 × 2C1 = 24 ways. Hence, the required answer is 18 + 24 = 42 ways. Hence, option (d) is correct.

How many numbers of 3-digits can be formed with the digits 1, 2, 3, 4, 5 (repetition of digits not allowed)?

Explanation :The number of numbers formed would be given by 5 × 4 × 3 (given that the first digit can be filled in 5 ways, the second in 4 ways and the third in 3 ways – MNP rule).

How many numbers between 2000 and 3000 can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7 (repetition of digits not allowed?

Explanation :The first digit can only be 2 (1 way), the second digit can be filled in 7 ways, the third in 6 ways and the fourth in 5 ways. A total of 1 × 7 × 6 × 5 = 210 ways.

In how many ways can a person send invitation cards to 6 of his friends if he has four servants to distribute the cards?

Explanation :Each invitation card can be sent in 4 ways. Thus, 4 × 4 × 4 × 4 × 4 × 4 = 46

In how many ways can 5 prizes be distributed to 8 students if each student can get any number of prizes?

Explanation :In this case since nothing is mentioned about whether the prizes are identical or distinct we can take the prizes to be distinct (the most logical thought given the situation). Thus, each prize can be given in 8 ways — thus a total of 85 ways.

In how many ways can 7 Indians, 5 Pakistanis and 6 Dutch be seated in a row so that all persons of the same nationality sit together?

Explanation :We need to assume that the 7 Indians are 1 person, so also for the 6 Dutch and the 5 Pakistanis. These 3 groups of people can be arranged amongst themselves in 3! ways. Also, within themselves the 7 Indians the 6 Dutch and the 5 Pakistanis can be arranged in 7!, 6! And 5! ways respectively. Thus, the answer is 3! × 7! × 6! × 5!.

There are 5 routes to go from Allahabad to Patna & 4 ways to go from Patna to Kolkata, then how many ways are possible for going from Allahabad to Kolkata via Patna?

Explanation :Use the MNP rule to get the answer as 5 × 4 = 20.

There are 4 qualifying examinations to enter into Oxford University: RAT, BAT, SAT, and PAT. An Engineer cannot go to Oxford University through BAT or SAT. A CA on the other hand can go to the Oxford University through the RAT, BAT & PAT but not through SAT. Further there are 3 ways to become a CA(viz., Foundation, Inter & Final). Find the ratio of number of ways in which an Engineer can make it to Oxford University to the number of ways a CA can make it to Oxford University

Explanation :An IITian can make it to IIMs in 2 ways, while a CA can make it through in 3 ways. Required ratio is 2:3.

How many straight lines can be formed from 8 non-collinear points on the X-Y plane?

Explanation :For a straight line we just need to select 2 points out of the 8 points available. 8C2 would be the number of ways of doing this.

Explanation :Use the property nCr = nCn-r to see that the two values would be equal at n = 11 since 11C3 = 11C8

In a chess tournament there were two women participating and every participant played two games with the other participants. The number of games that the men played among themselves exceeded the number of games that the men played with the women by 66. The number of participants in the tournament were?

Explanation :Solve this one through options. If you pick up option (a) it gives you 12 participants in the tournament. This means that there are 10 men and 2 women. In this case there would be 2 × 10C2 = 90 matches amongst the men and 2 × 10C1 × 2C1 = 40 matches between 1 man and 1 woman. The difference between number of matches where both participants are men and the number of matches where 1 participant is a man and one is a woman is 90 – 40 = 50 – which is not what is given in the problem. With 15 participants Æ 11 men and 2 women. In this case there would be 2 × 11C2 = 110 matches amongst the men and 2 × 11C1 × 2C1 = 44 matches between 1 man and 1 woman. The difference between number of matches where both participants are men and the number of matches where 1 participant is a man and one is a woman is 110 – 44 = 66 – which is the required value as given in the problem. Thus, option (b) is correct

The total number of games played in the tournament were ? Question same as above

Explanation :Based on the above thinking we get that since there are 15 players and each player plays each of the others twice, the number of games would be 2 × 15C2 = 2 × 105 = 210

How many even numbers of four digits can be formed with the digits 1, 2, 3, 4, 5, 6 (repetitions of digits are allowed)?

Explanation :Number of even numbers = 6 × 6 × 6 × 3 = 648

In how many ways five chocolates can be chosen from an unlimited number of Cadbury, Five-star, and Perk chocolates?

Explanation :For each selection there are 3 ways of doing it. Thus, there are a total of 3 × 3 × 3 × 3 × 3 = 243.

How many numbers can be formed with odd digits 1, 3, 5, 7, 9 without repetition?

Explanation :One digit no. = 5; Two digit nos = 5 × 4 = 20; Three digit no = 5 × 4 × 3 = 60; four digit no = 5 × 4 × 3 × 2 = 120; Five digit no.= 5 × 4 × 3 × 2 × 1 = 120 Total number of nos = 325

There are 6 boxes numbered 1, 2, ....6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

Explanation :With one green ball there would be six ways of doing this. With 2 green balls 5 ways, with 3 green balls 4 ways, with 4 green balls 3 ways, with 5 green balls 2 ways and with 6 green balls 1 way. So a total of 1 + 2 + 3 + 4 + 5 + 6 = 21 ways

Twenty seven persons attend a party. Which one of the following statements can never be true?

Explanation :Each of the first, third and fourth options can be obviously seen to be true— no mathematics needed there. Only the second option can never be true. In order to think about this mathematically and numerically— think of a party of 3 persons say A, B and C. In order for the second condition to be possible, each person must know a different number of persons. In a party with 3 persons this is possible only if the numbers are 0, 1 and 2. If A knows both B and C (2), B and C both would know at least 1 person— hence it would not be possible to create the person knowing 0 people. The same can be verified with a group of 4 persons i.e., the minute you were to make 1 person know 3 persons it would not be possible for anyone in the group to know 0 persons and hence you would not be able to meet the condition that every person knows a different number of persons

Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appear same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many four-letter computer passwords can be formed using only the symmetric letters (no repetition allowed)?

Explanation :Since there are 11 symmetric letters, the number of passwords that can be formed would be 11 × 10 × 9 × 8 = 7920.

Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appear same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three-letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

Explanation :This would be given by the number of passwords having: 1 symmetric and 2 asymmetric letters + 2 symmetric and 1 asymmetric letter + 3 symmetric and 0 asymmetric letters 11C1 × 15C2 × 3! + 11C2 × 15C1 × 3! + 11C3 × 3! = 11 × 105 × 6 + 55 × 15 × 6 + 11 × 10 × 9 = 6930 + 4950 + 990 = 12870

In how many ways is it possible to choose a white square and a black square on a chess board so that the squares must not lie in the same row or column?

Explanation :The white square can be selected in 32 ways and once the white square is selected 8 black squares become ineligible for selection. Hence, the black square can be selected in 24 ways. 32 × 24 = 768.

How many numbers greater than 0 and less than a million can be formed with the digits 0, 7 and 8?

Explanation :A million is 1000000 (i.e. the first seven digit number). So we need to find how many numbers of less than 7 digits can be formed using the digits 0,7 and 8. Number of 1 digit numbers = 2 Number of 2 digit numbers = 2 × 3 = 6 Number of 3 digit numbers = 2 × 3 × 3 = 18 Number of 4 digit numbers = 2 × 3 × 3 × 3 = 54 Number of 5 digit numbers = 2 × 3 × 3 × 3 × 3 = 162 Number of 6 digit numbers = 2 × 3 × 3 × 3 × 3 × 3 = 486 Total number of numbers = 728

Several teams take part in a competition, each of which must play one game with all the other teams. How many teams took part in the competition if they played 45 games in all?

Explanation :If the number of teams is n, then nC2 should be equal to 45. Trial and error gives us the value of n as 10.

In how many ways a selection can be made of at least one fruit out of 5 bananas, 4 mangoes and 4 almonds?

Explanation :From 5 bananas we have 6 choices available (0, 1, 2, 3, 4 or 5). Similarly 4 mangoes and 4 almonds can be chosen in 5 ways each. So total ways = 6 × 5 × 5 = 150 possible selections. But in this 150, there is one selection where no fruit is chosen. So required no. of ways = 150 – 1 = 149

There are 5 different Jeffrey Archer books, 3 different Sidney Sheldon books and 6 different John Grisham books. The number of ways in which at least one book can be given away is

Explanation :For each book we have two options, give or not give. Thus, we have a total of 2 14 ways in which the 14 books can be decided upon. Out of this, there would be 1 way in which no book would be given. Thus, the number of ways is 214 – 1

In the above problem, find the number of ways in which at least one book of each author can be given

Explanation :The number of ways in which at least 1 Archer book is given is (25–1). Similarly, for Sheldon and Grisham we have (23–1) and (26–1). Thus required answer would be the multiplication of the three (25–1)(23–1)(26–1)

There is a question paper consisting of 15 questions. Each question has an internal choice of 2 options. In how many ways can a student attempt one or more questions of the given fifteen questions in the paper

Explanation :For each question we have 3 choices of answering the question (2 internal choices + 1 nonattempt). Thus, there are a total of 315 ways of answering the question paper. Out of this there is exactly one way in which the student does not answer any question. Thus there are a total of 315 – 1 ways in which at least one question is answered

How many numbers can be formed with the digits 1, 6, 7, 8, 6, 1 so that the odd digits always occupy the odd places?

Explanation :The digits are 1, 6, 7, 8, 7, 6, 1. In this seven-digit no. there are four add places and three even places— OEOEOEO. The four odd digits 1, 7, 7, 1 can be arranged in four odd places in [4!/2! × 2] = 6 ways [as 1 and 7 are both occurring twice]. The even digits 6, 8, 6 can be arranged in three even places in 3!/2! = 3 ways. Total no. of ways = 6 × 3 = 18

There are five boys of McGraw-Hill Mindworkzz and three girls of I.I.M Lucknow who are sitting together to discuss a management problem at a round table. In how many ways can they sit around the table so that no two girls are together?

Explanation :We have no girls together, let us first arrange the 5 boys and after that we can arrange the girls in the spaces between the boys. Number of ways of arranging the boys around a circle = [5 – 1]! = 24. Number of ways of arranging the girls would be by placing them in the 5 spaces that are formed between the boys. This can be done in 5P3 ways = 60 ways. Total arrangements = 24 × 60 = 1440

Amita has three library cards and seven books of her interest in the library of Mindworkzz. Of these books she would not like to borrow the D.I. book, unless the Quants book is also borrowed. In how many ways can she take the three books to be borrowed?

Explanation :Books of interest = 7, books to be borrowed = 3 Case 1— Quants book is taken. Then D.I book can also be taken. So Amita is to take 2 more books out of 6 which she can do in 6C2 = 15 ways. Case 2— If Quants book has not been taken, the D.I book would also not be taken. So Amita will take three books out of 5 books. This can be done in 5C3 = 10 ways. So total ways = 15 + 10 = 25 ways.

From a group of 12 dancers, five have to be taken for a stage show. Among them Radha and Mohan decide either both of them would join or none of them would join. In how many ways can the 5 dancers be chosen

Explanation :We have to select 5 out of 12. If Radha and Mohan join- then we have to select only 5 – 2 = 3 dancers out of 12 – 2 = 10 which can be done in 10C3 = 120 ways. If Radha and Mohan do not join, then we have to select 5 out of 12 – 2 = 10-> 10C5 = 252 ways. Total number of ways = 120 + 252 = 372.

Find the number of 6-digit numbers that can be found using the digits 1, 2, 3, 4, 5, 6 once such that the 6-digit number is divisible by its unit digit. (The unit digit is not 1.)

Explanation :The unit digit can either be 2, 3, 4, 5 or 6. When the unit digit is 2, the number would be even and hence will be divisible by 2. Hence all numbers with unit digit 2 will be included which is equal to 5! Or 120. When the unit digit is 3, then in every case the sum of the digits of the number would be 21 which is a multiple of 3. Hence all numbers with unit digit 3 will be divisible by 3 and hence will be included. Total number of such numbers is 5! or 120. Similarly for unit digit 5 and 6, the number of required numbers is 120 each. When the unit digit is 4, then the number would be divisible by 4 only if the ten’s digit is 2 or 6. Total number of such numbers is 2 × 4! or 48. Hence total number of required numbers is (4 × 120) + 48 = 528.

An urn contains 5 boxes. Each box contains 5 balls of different colours red, yellow, white, blue and black. Rangeela wants to pick up 5 balls of different colours, a different coloured ball from each box. If from the first box in the first draw, he has drawn a red ball and from the second box he has drawn a black ball, find the maximum number of trials that are needed to be made by Rangeela to accomplish his task if a ball picked is not replaced.

Explanation :As we need to find the maximum number of trials, so we have to assume that the required ball in every box is picked as late as possible. So in the third box, first two balls will be red and black. Hence third trial will give him the required ball. Similarly, in fourth box, he will get the required ball in fourth trial and in the fifth box, he will get the required ball in fifth trial. Hence maximum total number of trials required is 3 + 4 + 5 = 12.

How many rounds of matches does a knock-out tennis tournament have if it starts with 64 players and every player needs to win 1 match to move at the next round?

Explanation :Since every player needs to win only 1 match to move to the next round, therefore the 1st round would have 32 matches between 64 players out of which 32 will be knocked out of the tournament and 32 will be moved to the next round. Similarly in 2nd round 16 matches will be played, in the 3rd round 8 matches will be played, in 4th round 4 matches, in 5th round 2 matches and the 6th round will be the final match. Hence total number of rounds will be 6 (26 = 64).

There are N men sitting around a circular table at N distinct points. Every possible pair of men except the ones sitting adjacent to each other sings a 2 minute song one pair after other. If the total time taken is 88 minutes, then what is the value of N?

Explanation :Total number of pairs of men that can be selected if the adjacent ones are also selected is NC2. Total number of pairs of men selected if only the adjacent ones are selected is N. Hence total number of pairs of men selected if the adjacent ones are not selected is NC2 – N. Since the total time taken is 88 min, hence the number of pairs is 44. Hence, NC2 – N = 44 so N = 11

Q. Number of ways of arranging MALAYALAM such that vowels are never together.

A. Total ways = 5 letters and 4 vowels; Vowels can be treated as single entity giving us 6 alphabets; So 6P6 = 6! but M and L are repeated twice so removing duplicate entries we get 6!/(2!*2!) ways.

Total ways in which 9 letters can be arranged are = 9!/(2!*2!*4!) ways removing duplicates for 2 M's, 2 L's and 4 A's

Total ways vowels are never together = Total ways - Total ways were they are together

Q. How many ways can a cricket team of 11 be chosen out of batch of 15 players.

A. ways = 15C11 = 15C4

Q. How many ways to select a committee of 3 men and 2 ladies from 6 men and 5 ladies.

A. Men can be chosen in 6C3 ways and women in 5C2 ways.

Total ways are = 6C3 * 5C2.

Q. There are 6 boxes numbered 1,2,… 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

Ans . B

GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR, RRRRRG GGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGG GGGRRR, RGGGRR, RRGGGR, RRRGGG GGGGRR, RGGGGR, RRGGGG GGGGGR, RGGGGG GGGGGG
Hence 21 ways.

Q. Let T be the set of integers {3, 11, 19, 27, …, 451, 459, 467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

Ans . D

Tn = a + (n-1)d
467 = 3 + (n – 1)8 ⇒ n = 59
Half of n = 29 terms 29th term is 227 and 30 th term is 235 and when these two terms are added the sum is less than 470. Hence the maximum possible values the set S can have is 30.

Q. The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n - 1)(n - 2).. 3.2.1 is not divisible by n is

Ans . B

From 12 to 40, there are 7 prime numbers, i.e., 13, 17, 19, 23, 29, 31 and 37 such that (n – 1)! is not divisible by any of them.

Q. A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition

11 ≤ e ≤ 66
10 ≤ e ≤ 66
11 ≤ e ≤ 65
0 ≤ e ≤ 11

Ans . A

The least number of edges will be when one point is connected to each of the other 11 points, giving a total of 11 lines. One can move from any point to any other point via the common point. The maximum edges will be when a line exists between any two points. Two points can be selected from 12 points in
12C2i.e. 66 lines

Q. In a certain examination paper, there are n questions. For j = 1,2 …n, there are 2n-j students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is

Ans . A

Let us say there are only 3 questions. Then, there are
23-1 = 4 students who have done 1 or more questions wrongly, 23-2 = 2 students who have done 2 or more questions wrongly and 23-3 = 1 student who must have done all 3 wrongly. Thus, total number of wrong answers = 4 + 2 + 1 = 7 = 23 – 1 = 2n – 1.
= 4095 = 212 – 1. Thus n = 12