In how many ways can be the letter of the word ‘STRANGE’ be arranged so that(a) The vowel may appear in the odd places(b) The vowels are never separated(c) The vowels never come together
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
ways
And 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.
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GMAT Club Legend Joined: 08 Jul 2010 Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator Posts: 5912 Location: India GMAT: QUANT EXPERT WE:Education (Education)
In how many ways can the letters of word "EDUCATION" be arranged such [#permalink]
Updated on: 08 Jul 2015, 07:04
00:00 Question Stats: 78%
(01:38) correct 22% (01:50) wrong based on 114 sessions Hide Show timer Statistics 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! _________________ GMATinsight
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Originally posted by GMATinsight on 08 Jul 2015, 06:41.
Last edited by Bunuel on 08 Jul 2015, 07:04, edited 1 time in total. Renamed the topic and edited the question. SVP Joined: 20 Mar 2014 Posts: 2417 Concentration: Finance, Strategy GMAT 1: 750 Q49 V44 GPA: 3.7 WE:Engineering (Aerospace and Defense) Re: In how many ways can the
letters of word "EDUCATION" be arranged such [#permalink]
08 Jul 2015, 06:50 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.
_________________
Senior Manager Joined: 15 Sep 2011 Posts: 281 Location: United States WE:Corporate Finance (Manufacturing) In how many ways can the letters of word "EDUCATION" be arranged such
[#permalink] 08 Jul 2015, 10:30 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, Intern Joined: 05 Nov 2014 Posts: 38 Concentration: Marketing, International Business Re: In how many ways can the letters of word "EDUCATION" be arranged such
[#permalink] 26 Jul 2015, 22:52 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 GMAT Club Legend Joined: 08 Jul 2010 Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator Posts: 5912 Location: India GMAT: QUANT EXPERT WE:Education (Education) In how many ways can the letters of word "EDUCATION" be arranged such [#permalink]
26 Jul 2015, 23:24 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 U
and 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!
ways
i.e. Total Arrangements = 5!*4!
Answer: option B
_________________ GMATinsight
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Sol. l NEW:QUANT REVISION Topicwise Quiz Intern Joined: 05 Nov 2014 Posts: 38 Concentration: Marketing, International Business In how many ways can the letters of word "EDUCATION" be arranged such
[#permalink] 27 Jul 2015, 04:37 Quote: 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 U
and 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! ways
i.e. Total Arrangements = 5!*4!
Answer: option B
Sorry! I misread the question! I need to pay more attention to detail.. Nice explanation btw! Non-Human
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Joined: 09 Sep 2013 Posts: 24692 Re: In how many ways can the letters of word "EDUCATION" be arranged such [#permalink]
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_________________ Re: In how many ways can the letters of word "EDUCATION" be arranged such [#permalink] 01 Sep 2021, 18:34
Moderators: Senior Moderator - Masters Forum 3088 posts
How many ways can one arrange the word EDUCATION such that relative positions of vowels and consonants remains same?
Hence, the total number of ways = 4!
How many ways can arrange the word EDUCATION such that all vowels come together?
There are three possible vowel clusters, corresponding to the position of the e among the a's. Considering the vowel cluster as a single letter, there are 5 letters, and therefore 5!= 120 ways to arrange them.
How many ways are there to arrange the letters of the word EDUCATION so that?
There are fifteen solutions to this reduced equation.
How many ways can one arrange EDUCATION?
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!*
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