You have not accounted for those arrangements such as EAUDCTOIN in which there are disjoint pairs of adjacent vowels. What we need to exclude are adjacent pairs of vowels. Notice that we can partition $5$ in the following ways: \begin{align*} 5 & = 4 + 1\\ & = 3 + 2\\ & = 3 + 1 + 1\\ & = 2 + 2 + 1\\ & = 2 + 1 + 1 + 1\\ & = 1 + 1 + 1 + 1 + 1 \end{align*} Think of these numbers as the number of consecutive vowels. The case $1 + 1 + 1 + 1 + 1$ in which no two vowels are adjacent is what we want to count. The case $2 + 1 + 1 + 1$ has one pair of adjacent vowels. The case $2 + 2 + 1$ has two disjoint pairs of adjacent vowels. The case $3 + 1 + 1 + 1$ has two overlapping pairs of adjacent vowels (such as the string AEI, which has the pairs AE and EI). The case $3 + 2$ has three pairs of adjacent vowels, two of which are overlapping. The case $4 + 1$ has three pairs of adjacent vowels. The case $5$ has four pairs of adjacent vowels. We use the Incluson-Exclusion Principle to count the number of admissible choices for the positions of the vowels and consonants first, then arrange the vowels and consonants in the those positions. There are $\binom{9}{5}$ ways to choose the positions of the vowels. From these, we must exclude those arrangements in which there are one or more pairs of adjacent vowels. A pair of adjacent vowels: We have eight positions, one for a block of two vowels, three for single vowels, and four for consonants. Choose one position for the block and three of the remaining seven positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are $$\binom{8}{1}\binom{7}{3}$$ such choices. Two pairs of adjacent vowels: This can occur in two ways. Either there are two overlapping pairs (a block of three consecutive vowels) or two disjoint pairs (two blocks of two vowels each). Two overlapping pairs: We have seven positions to fill with a block of three vowels, two single vowels, and four consonants. Choose one position for the block and two of the remaining six positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are $$\binom{7}{1}\binom{6}{2}$$ such choices. Two disjoint pairs: We have seven positions to fill with two blocks of two vowels, one single vowel, and four consonants. Choose two positions for the blocks and one of the remaining five positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are $$\binom{7}{2}\binom{5}{1}$$ such choices. Three pairs of adjacent vowels: There are again two cases. Either there are three overlapping pairs of adjacent vowels (a block of four consecutive vowels) or two overlapping pairs of vowels (a block of three consecutive vowels) and a disjoint pair of adjacent vowels (a block of two consecutive vowels). Three overlapping pairs of adjacent vowels: There are six positions to fill with a block of four consecutive vowels, a single vowel, and four consonants. There are six ways to choose the position of the block and five ways to choose the position of the single vowel. The remaining four positions must be reserved for the four consonants. There are $$\binom{6}{1}\binom{5}{1}$$ such choices. Two overlapping pairs of adjacent vowels and a disjoint pair of adjacent vowels: There are six positions to fill with a block of three consecutive vowels, a block of two consecutive vowels, and four consonants. There are six ways to choose the position of the block of three vowels and five ways to choose the position of the block of two vowels. The remaining four positions must be reserved for the four consonants. There are $$\binom{6}{1}\binom{5}{1}$$ such choices. Four pairs of adjacent vowels: There are five positions to fill with a block of five consecutive vowels and four consonants. Choose the position of the block. The remaining four positions must be reserved for the four consonants. There are $$\binom{5}{1}$$ such choices. By the Inclusion-Exclusion Principle, the number of ways of positioning the vowels and consonants so that no two vowels are consecutive is $$\binom{9}{5} - \binom{8}{1}\binom{7}{3} + \binom{7}{1}\binom{6}{2} + \binom{7}{2}\binom{5}{1} - \binom{6}{1}\binom{5}{1} - \binom{6}{1}\binom{5}{1} + \binom{5}{1} = 1$$ namely, VCVCVCVCV. There are $5!$ ways to arrange the vowels in their five positions and $4!$ ways to arrange the consonants in their four positions. Hence, the number of admissible arrangements is $$\left[\binom{9}{5} - \binom{8}{1}\binom{7}{3} + \binom{7}{1}\binom{6}{2} + \binom{7}{2}\binom{5}{1} - \binom{6}{1}\binom{5}{1} - \binom{6}{1}\binom{5}{1} + \binom{5}{1}\right]5!4! = 5!4!$$ which agrees with your answer.
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In how many ways can the letters of word "EDUCATION" be arranged such [#permalink]
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Question Stats: Hide Show timer StatisticsIn how many ways can the letters of word "EDUCATION" be arranged such that NO two vowels appear together?A) 9!B) 5!*4!C) 5!*5!D) 5!*4!*2! E) 6!*4! _________________
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Originally posted by GMATinsight on 08 Jul 2015, 06:41. Renamed the topic and edited the question.
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Re: In how many ways can the letters of word "EDUCATION" be arranged such [#permalink]
GMATinsight wrote: In how many ways can the letters of word "EDUCATION" be arranged such that NO two vowels appear together?A) 9!B) 5!*4!C) 5!*5!D) 5!*4!*2! E) 6!*4! No 2 vowels together = the only arrangement possible will be V C V C V C V C V (with V=vowel, C=consonant). This is true as we have 5 vowels and 4 consonants and any other combination will force us to pair 2 vowels together.Thus, the number of arrangements possible : 5 *4 *4 *3 *3 *2 *2*1 = 5!*4! ----> B is the correct answer. _________________
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In how many ways can the letters of word "EDUCATION" be arranged such [#permalink] There are five vowels and four consontants. The consontants must be placed in between each vowel in order to fulfill the requirement that no two vowels touch each other. Thus, answer choice B. \(5! * 4!\). Answer choice A, C, D, and E don't make sense. A is for when all choices are distinct, but in fact half of them are, vowels or consonants. Answer choices C and E include another integer 5 and 6, respectively, for no reason at all, thereby making the number of selections 10 instead of nine. As well, answer choice D appears as though the selections had to be doubled to count for whether consontants or vowels start first, but the factorial itself accounts for it and none of the consonants could begin first because of the constraint. So nothing of such is warranted. Thanks,
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Re: In how many ways can the letters of word "EDUCATION" be arranged such [#permalink]
Engr2012 wrote: GMATinsight wrote: In how many ways can the letters of word "EDUCATION" be arranged such that NO two vowels appear together?A) 9!B) 5!*4!C) 5!*5!D) 5!*4!*2! E) 6!*4! No 2 vowels together = the only arrangement possible will be V C V C V C V C V (with V=vowel, C=consonant). This is true as we have 5 vowels and 4 consonants and any other combination will force us to pair 2 vowels together.Thus, the number of arrangements possible : 5 *4 *4 *3 *3 *2 *2*1 = 5!*4! ----> B is the correct answer. We can consider the consonants as one group: the set looks like [D,C,T,N], E, U, A, I, O where the [] is regarded as one item.Number of ways we can arrange the set = 6! (since there are 6 items)Number of ways we can arrange [D,C,T,N] = 4!Required arrangement = 6! * 4! OA is E
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In how many ways can the letters of word "EDUCATION" be arranged such [#permalink]
Subanta wrote: Engr2012 wrote: GMATinsight wrote: In how many ways can the letters of word "EDUCATION" be arranged such that NO two vowels appear together?A) 9!B) 5!*4!C) 5!*5!D) 5!*4!*2! E) 6!*4! No 2 vowels together = the only arrangement possible will be V C V C V C V C V (with V=vowel, C=consonant). This is true as we have 5 vowels and 4 consonants and any other combination will force us to pair 2 vowels together.Thus, the number of arrangements possible : 5 *4 *4 *3 *3 *2 *2*1 = 5!*4! ----> B is the correct answer. We can consider the consonants as one group: the set looks like [D,C,T,N], E, U, A, I, O where the [] is regarded as one item.Number of ways we can arrange the set = 6! (since there are 6 items)Number of ways we can arrange [D,C,T,N] = 4!Required arrangement = 6! * 4! OA is E Hi Subanta You seem to have read the question wrong. Question says that "such that NO two vowels appear together?" EDUCATION has 5 Vowels A E I O Uand all must be separated by atleast one consonant between any two adjacent Vowels and we only have 4 consonants DCTN i.e. in the arrangemtn A - E - I - O - U all teh dashes (-) must be occupied by atleast 1 consonant and since we have 4 places and 4 consonants so every dash (-) must have exactly one consonant.i.e. the arrangement will be A D E C I T O N U where all vowels AEIOU can exchange positions among themselves in 5! ways and similarly all Consonants DCTN can exchange positions among themselves in 4! waysi.e. Total Arrangements = 5!*4!Answer: option B _________________
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In how many ways can the letters of word "EDUCATION" be arranged such [#permalink]
Quote:
You seem to have read the question wrong. Question says that "such that NO two vowels appear together?" EDUCATION has 5 Vowels A E I O Uand all must be separated by atleast one consonant between any two adjacent Vowels and we only have 4 consonants DCTN i.e. in the arrangemtn A - E - I - O - U all teh dashes (-) must be occupied by atleast 1 consonant and since we have 4 places and 4 consonants so every dash (-) must have exactly one consonant.i.e. the arrangement will be A D E C I T O N U where all vowels AEIOU can exchange positions among themselves in 5! ways and similarly all Consonants DCTN can exchange positions among themselves in 4! waysi.e. Total Arrangements = 5!*4! Answer: option B Sorry! I misread the question! I need to pay more attention to detail.. Nice explanation btw!
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Re: In how many ways can the letters of word "EDUCATION" be arranged such [#permalink] Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: In how many ways can the letters of word "EDUCATION" be arranged such [#permalink] 01 Sep 2021, 18:34 |
How many different words can be formed with the letters of the word EDUCATION such that all vowels are together?
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