In how many ways can the letters of the word permutations be arranged

Question

In how many ways can the letters of the word permutations be arranged such that the order of vowels remains unchanged?

A.  \frac{12!}{2!}

B.  \frac{12!}{5!}

C.  \frac{12!}{2!5!}

D.  \frac{4*7!}{2! }

E.  \frac{4*7!}{2! 5!}

M37-44

Bunuel · Accepted Answer

Official Solution:

In how many ways can the letters of the word PERMUTATIONS be arranged such that the order of vowels remains unchanged?

A.  \frac{12!}{2!} 
B.  \frac{12!}{5!} 
C.  \frac{12!}{2!*5!} 
D.  \frac{4*7!}{2!} 
E.  \frac{4*7!}{2!*5!}

PERMUTATIONS has 12 letters: 5 vowels (EUAIO) and 7 consonants (PRMTTNS).
 
Choose 5 slots from 12 for the vowels:  C^5_{12}=\frac{12!}{5!*7!} . Since the order of the vowels must remain unchanged (EUAIO), then we do not arrange them.
 
7 consonants in 7 remaining slots can be arranged in  \frac{7!}{2!}  ways (notice that we have two T's among 7 consonants).
 
So, the total number of ways is  \frac{12!}{5!*7!}*\frac{7!}{2!}=\frac{12!}{2!*5!} .

Answer: C

GMATinsight · Answer

permutations has 5 vowels and 7 consonants

total ways of choosing places out of 7 for 12 = 12C5  = \frac{12!}{5!*7!}

Arrangement of remaining 7 consonants on 7 empty places  = \frac{7!}{2!}  (because 2 of these 7 are identical 't')

Total arrangemnts  = \frac{12!}{5!*7!} * \frac{7!}{2!} = \frac{12!}{5!*2!}

Answer: Option C

sthahvi · Answer

Wouldn't ways of choosing 7 places out of 12 be 12C7 also why do we need to choose the places as the positions of the vowels are fixed, so there is an option from only 7!/2! places. how do we need to choose the places ?

JoaquinCerda · Answer

I have the same understanding because we volwels stay the same so we should not consider them.

What changes are the consonats which are 7 (with 2 "T" "T)
meaning 7!/2!

Bambi2021 · Answer

If the places of the vowels stay the same, then I think we can treat them as identical.

PERMUTATIONS = PERMETETEENS

However, we must still include them in the formula because we can have different numbers of consonants between them. We can have

PRMTTNSEEEEE = PRMTTNSEUAIO

or we can have

PRMTEEETEENS = PRMTEUATIONS

If we only arrange the consonants, as is done by 7!, then we fail to account for this.

So in total we have 12 letters with 5 identical vowels and 2 identical consonants.

Posted from my mobile device

Regor60 · Answer

12 letters can be arranged:

12! ways

But only one of the 5! arrangements of the 5 vowels counts:

12!/5!

Finally, since there are two "t" letters:

12!/5!2!

Posted from my mobile device

Dream009 · Answer

my thinking was 12 characters with t in common can be arranged in 12!/2! and then we divide by 5! coz the vowels can arrange themselves in that many ways and we want only 1 order of them.

In how many ways can the letters of the word permutations be arranged

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