In how many different ways can the letters of the word "CORPORATION"

There are several problems with this approach:
1. The number of arrangements of CORPORATION is 11!/(2!*3!) (we should divide by 2! and 3! because of 2 R's and 3 O's there);

2. When you consider the vowels as one units as {OOAIO}, you get 7 units {OOAIO}, {C}, {R}, {P}, {R}, {T}, {N}. The number of arrangements of these 7 units is 7!/2! (again we divide by 2! because of 2 R's). OOAIO within their unit can also be arranged in 5!/3! ways (5 letters with three O's). So, the number of arrangements when vowels are together is 7!/2!*5!/3!.

3. When you subtract the cases when the vowels are together from total ways you don't get the cases where no two vowels are together, you just get the cases when ALL 5 vowels are not together, which is not what we wanted. For example, one of the cases you get would be CORPORATION, where, as you can see, I and O are together.

So, the way you are doing is not a good way to approach this problem.

THEORY

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\)

For example number of permutation of the letters of the word "gmatclub" is \(8!\) as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Hope this helps.

Thank you, your explanation is crystal clear, and your ability to solve pretty much any gmat math question is astonishing!
Thank you for taking the time out to answer my queries.

11! - ( 7! X 5) is the answer

CORPORATION

C - Used 1 time
O - Used 3 times
R - Used 2 times
P - Used 1 time
A - Used 1 time
T - Used 1 time
I - Used 1 time
N - Used 1 time

Total Letters = 11

Total Vowels (A, E, I, O, U) = 5

Requirement = Vowels anywhere adjacent to consonants C C C C C C

We have 7 places available to vowels but we need only 5

So Total Ways for vowels = 7C5*5!/3!

Total Ways for consonants = 6!/2!

Total POssible arrangements as reqquired = (7C5*5!/3!)*(6!/2!)

I don't understand Step 3. Please explain. We have 5 vowels and 5 places, why are we choosing 5 out of 7 slots for them?

No two vowels must be together. If 5 vowels take places of *'s in *C*R*P*R*T*N*, then no two vowels will be together. There are 7 *'s so the number of ways we can do this is \(C^5_7\).

Hope it's clear.

I am a bit confused here.We have to select 5 out of 7 blank spaces or '*' in this case.Why will it be \(C^5_7\) . Should it not be the other way round as in 7C5?
Thanks

\(C^5_7\), \(C^7_5\), 7C5 are the same: choosing 5 out of 7. How can we choose 7 out of 5? Those are just different ways of writing the same.

(11!/2!*3!)- (5!*7!/3!*2!)
Total ways of rearranging the word is 11!/2!*3! As r repeats 2 time and O repeats 3 times. We assume all vowels as 1 block(they will always be together) and it is arranged with all other letters we get 7!/2! And the vowels can be arranged amongst themselves in 5!/3! ways. So total number of ways vowels can never be together would be
(11!/2!*3!)- (5!*7!/3!*2!)
Bunuel please correct me if I am wrong

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Bunuel Would this be an equally valid method?

11! / 3! x 2! <--- Perms with repeats

Options:
a) All the vowels are together : 7! / 2!
b) 4 together: 8! / 2!
c) 3 together: 9! / 2!
d) 2 together: 10! / 2!

11! / 3! x 2! - (7! / 2!) - (8! / 2!) - (9! / 2!) - (10! / 2!)

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