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The number of arrangement of letters of the word BANANA in [#permalink]
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Question Stats: Hide Show timer StatisticsThe number of arrangement of letters of the word BANANA in which the two N's do not appear adjacently is:A. 40B. 50C. 60D. 80 E. 100
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Re: The number of arrangement of letters of the word BANANA in [#permalink]
saurabhprashar wrote: The number of arrangement of letters of the word BANANA in which the two N's do not appear adjacently is:A. 40B. 50C. 60D. 80 E. 100 Total # of arrangements of BANANA is 6!/(3!2!) = 60 (arrangement of 6 letters {B}, {A}, {N}, {A}, {N}, {A}, where 3 A's and 2 N's are identical).The # of arrangements in which the two N's ARE together is 5!/3!=20 (arrangement of 5 units letters {B}, {A}, {A}, {A}, {NN}, where 3 A's are identical)..The # of arrangements in which the two N's do not appear adjacently is {total} - {restriction} = 60 - 20 = 40.Answer: A. _________________
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The number of arrangement of letters of the word BANANA in [#permalink] BunuelThese types of multinomial questions are great.Is there a name for when two objects can't be adjacent to each other in combinations/permutations? It's a rule that I always forget how to apply. A long shot, but do you know the source for OPs question?
Re: The number of arrangement of letters of the word BANANA in [#permalink]
philipssonicare wrote: BunuelThese types of multinomial questions are great.Is there a name for when two objects can't be adjacent to each other in combinations/permutations? It's a rule that I always forget how to apply. A long shot, but do you know the source for OPs question? I call these questions MISSISSIPPI questions. To see why, see my post below Cheers,Brent _________________
Brent Hanneson – Creator of gmatprepnow.comI’ve spent the last 20 years helping students overcome their difficulties with GMAT math, and the biggest thing I’ve learned is… Many students fail to maximize their quant score NOT because they lack the skills to solve certain questions but because they don’t understand what the GMAT is truly testing - Learn more
Re: The number of arrangement of letters of the word BANANA in [#permalink]
saurabhprashar wrote: The number of arrangement of letters of the word BANANA in which the two N's do not appear adjacently is:A. 40B. 50C. 60D. 80 E. 100 ---------ASIDE-------------When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....] So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:There are 11 letters in total There are 4 identical I's There are 4 identical S's There are 2 identical P's So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]-------ONTO THE QUESTION!!---------------------------------In BANANA, there are: There are 6 letters in total There are 3 identical A's There are 2 identical N's So, the total number of possible arrangements = 6!/[(3!)(2!)] = 60 IMPORTANT: Among these 60 outcomes, there are some outcomes that break the rule about N's not appearing next to each other. So, let's determine the number of outcomes in which the 2 N's ARE together, and we'll subtract this from our 60 outcomes. First "glue" the 2 N's together, to get one "super letter" NN So, we now must arrange 5 letters: B, A, A, A, and NNThere are 5 letters in total There are 3 identical A's So, the total number of possible arrangements = 5!/(3!) = 20 Number of arrangements that adhere to the rule = 60 - 20 = 40 Answer: ACheers,Brent _________________
Brent Hanneson – Creator of gmatprepnow.comI’ve spent the last 20 years helping students overcome their difficulties with GMAT math, and the biggest thing I’ve learned is… Many students fail to maximize their quant score NOT because they lack the skills to solve certain questions but because they don’t understand what the GMAT is truly testing - Learn more
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Re: The number of arrangement of letters of the word BANANA in [#permalink] Hi Bunuel, Could you please confirm that while calculating restriction why we do not take into account reciprocal of {N1N2} i.e.{N2N1}? Is that because the reciprocal/repetition has already been excluded from the total combinations (by dividing 2!)? Thanks.
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Re: The number of arrangement of letters of the word BANANA in [#permalink] Hi Bunuel Could you please confirm that while calculating restriction why we do not take into account reciprocal of {N1N2} i.e.{N2N1}? Is that because the reciprocal/repetition has already been excluded from the total combinations (by dividing 2!)? Thanks.
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Re: The number of arrangement of letters of the word BANANA in [#permalink]
saurabhprashar wrote: The number of arrangement of letters of the word BANANA in which the two N's do not appear adjacently is:A. 40B. 50C. 60D. 80 E. 100 BANANA =arrangement = 6!/2!*3!= 60 waysand NNAAABNN=XXAAAB = 5!/3! = 20 waystotaal N is together ; 60-20 ; 40 waysIMO A
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Re: The number of arrangement of letters of the word BANANA in [#permalink] Total number of ways to arrange the letters of the given word BANANA : \(\frac{(6!) }{ (3! * 2!)}\) = 60Number of ways if both NN are adjacent: Consider 'NN' as 1 letter: BAAA (NN)=> We have 5 letters to arrange, therefore: 5!. => Both 'NN' can be arranged it 2! ways. As we have repetition (3A's and 2N's) therefore:=> \(\frac{(5!) *(2!) }{ (3! * 2!)}\) = 20The number of ways 'NN' won't be adjacent to each other: 60 - 20 = 40. Answer A _________________
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Re: The number of arrangement of letters of the word BANANA in [#permalink]
YassiHASHMI wrote: Hi Bunuel Could you please confirm that while calculating restriction why we do not take into account reciprocal of {N1N2} i.e.{N2N1}? Is that because the reciprocal/repetition has already been excluded from the total combinations (by dividing 2!)? Thanks. Hope it is clear.
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Re: The number of arrangement of letters of the word BANANA in [#permalink] 6! / 3! x 2! = 60 <--- Total # of permutations with repeats5! / 3! = 20 <--- Total # of permutations where Ns are TOGETHER60 - 20 = 40 <--- Ps where they are not togetherAnswer is A. _________________
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Re: The number of arrangement of letters of the word BANANA in [#permalink] arrangement = 6!/2!*3!= 60 waysand NNAAABNN=XXAAAB = 5!/3! = 20 waystotaal N is together ; 60-20 ; 40 ways Posted from my mobile device
Re: The number of arrangement of letters of the word BANANA in [#permalink] 15 Aug 2022, 08:19 |
What is the number of ways of arranging the letters of the word BANANA so that no two ns appear?
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