 Answer Hint:There are 7 letters in the word ‘STRANGE’ in which two are vowels. There are four odd places. So, we have to find the total number of ways for these four places. The other 5 letters may be placed in the five places in 5! ways. By this we get the total number of required arrangements. If vowels are never separated, in this case consider all the vowels a single letter. If all the vowels never come together in this case, we subtract all vowels together from the total number of arrangements. Complete step-by-step answer: For finding number of ways of placing n things in r objects, we use ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Also, for arranging n objects without restrictions, we have $n!$ ways.(a) There are 7 letters in the word ‘STRANGE’ amongst which 2 are vowels and there are 4 odd places (1, 3, 5, 7) where these two vowels are to be placed together.Total number of ways can be expressed by replacing n = 4 and r = 2,$\begin{align} & {}^{4}{{P}_{2}}=\dfrac{4!}{\left( 4-2 \right)!} \\ & =\dfrac{4!}{2!} \\ & =\dfrac{4\times 3\times 2!}{2!} \\ & =4\times 3=12 \\ \end{align}$And corresponding to these 12 ways the other 5 letters may be placed in 5! ways $=5\times 4\times 3\times 2\times 1=120$.Therefore, the total number of required arrangements = $12\times 120=1440$ ways.(b) Now, vowels are not to be separated. So, we consider all vowels as a single letter.There are six letters S, T, R, N, G, (AE) , so they can arrange themselves in 6! ways and two vowels can arrange themselves in 2! ways.Total number of required arrangements \[=6!\times 2!=6\times 5\times 4\times 3\times 2\times 1\times 2\times 1=1440\] ways.(c) Now, for the number of arrangements when all vowels never come together, we subtract the total arrangement where all vowels have occurred together from the total number of arrangements.The total number of arrangements = 7! = 5040 waysAnd the number of arrangements in which the vowels do not come together $=7!-6!2!$number of arrangements in which the vowels do not come together $=5040 -1440 = 3600$ ways.Note: Knowledge of permutation and arrangement of objects under some restriction and without restriction is must for solving this problem. Students must be careful while making the possible cases. They must consider all the permutations in word like in part (b), a common mistake is done by not considering the arrangements of vowels.
Answer: c
This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Permutations and Combinations”. 1. If 16Pr-1 : 15Pr-1 = 16 : 7 then find r. a) 10 b) 12 c) 7 d) 8 View Answer Answer: a Explanation: We know that \(^nP_r = \frac{n!}{(n-r)!} \) Hence \(^{16}P_{r-1} : \, ^{15}P_{r-1} = 16 : 7 \) \([\frac{16!}{16-(r-1)!}] ÷ [\frac{15!}{15-(r-1)}] = 16 ÷ 7 \) 16 ÷ (17 – r) = 16 ÷ 7 17 – r = 7 Hence r = 10. 2. In a colony, there are 55 members. Every member posts a greeting card to all the members. How many greeting cards were posted by them? a) 990 b) 890 c) 2970 d) 1980 View Answer Answer: c Explanation: First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players. Similarly, it happens with the rest of the players. The total numbers of greeting cards posted are 54 + 54 + 54 … 54 (55times) = 54 x 55 = 2970. 3. Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place. a) 36 b) 48 c) 144 d) 96 View Answer Answer: c 4. Find the sum of all four digit numbers that can be formed by the digits 1, 3, 5, 7, 9 without repetition. a) 666700 b) 666600 c) 678860 d) 665500 View Answer Answer: b Explanation: The given digits are 1, 3, 5, 7, 9 Sum of r digit number= n-1Pr-1 (Sum of all n digits)×(1111… r times) N is the number of non zero digits. Here n=5, r=4 The sum of 4 digit numbers4P3 (1+3+5+7+9)(1111)=666600. 5. If nPr = 3024 and nCr = 126 then find n and r. a) 9, 4 b) 10, 3 c) 12, 4 d) 11, 4 View Answer Answer: a Explanation: \(\frac{^nP_r}{^nC_r} = \frac{3024}{126} \) \(^nP_r = \frac{n!}{(n-r)!} \) \(^nC_r = \frac{n!}{(n-r)!×r!} \) Hence \( [\frac{n!}{(n-r)!}]÷[\frac{n!}{(n-r)!×r!}] \) = 24 24 = r! Hence r = 4 Now nP4 = 3024 \(\frac{n!}{(n-4)!} = 3024 \) n(n-1)(n-2)(n-3) = 9.8.7.6n = 9.
Check this: Engineering Mathematics MCQ | Numerical Methods MCQ 6. Find the number of rectangles and squares in an 8 by 8 chess board respectively. a) 296, 204 b) 1092, 204 c) 204, 1092 d) 204, 1296 View Answer Answer: b 7. There are 20 points in a plane, how many triangles can be formed by these points if 5 are colinear? a) 1130 b) 550 c) 1129 d) 1140 View Answer Answer: a Explanation: Number of points in plane n = 20. Number of colinear points m = 5. Number of triangles from by joining n points of which m are colinear = nC3 – mC3 Therefore the number of triangles = 20C3 – 5C3 = 1140-10 = 1130. 8. In how many ways can we select 6 people out of 10, of which a particular person is not included? Answer: c 9. Number of circular permutations of different things taken all at a time is n!. a) True b) False View Answer Answer: b 10. Is the given statement true or false? View Answer Answer: a \(^nC_{n-r} = \frac{n!}{r!×(n-r)!} = ^nC_r. \) Sanfoundry Global Education & Learning Series – Probability and Statistics. To practice all areas of Probability and Statistics, here is complete set of 1000+ Multiple Choice Questions and Answers.
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How many different words can be formed using all the letters of the word danger when no vowel occupies odd place?
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