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

Question

In how many different ways can the letters of the word "CORPORATION" be arranged in such a way that no two vowels are together?

Sorry I don't have an answer to this

Bunuel · Accepted Answer

Step 1: Separate consonants with 7 empty slots *C*R*P*R*T*N*.

Step 2: 6 consonants CRPRTN can be arranged in 6!/2! ways (6 letters with two R's);

Step 3: we have 5 vowels OOAIO. Choose 5 slots out of 7 for them (this way no two vowels will be together):  C^5_7 .

Step 4: 5 vowels in their slots can be arranged in 5!/3! ways (5 letters with three O's).

Total =  \frac{6!}{2!}*C^5_7*\frac{5!}{3!} .

Bunuel · Answer

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 CORPORAT IO N, 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.

usre123 · Answer

Thank you so much!

I just wanted to know, there is another way such questions are done: total ways these can be arranged (11!) minus the number of ways in which the vowels are together CRPRTNX where X is all the vowels bunched together. so that's 7!.
So we can have 11!-7!. But I'm not clear where to incorporate the multiple Os and Rs.
Or is this concept completely wrong?

usre123 · Answer

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.

arjunperiyasamys · Answer

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

GMATinsight · 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!)

profileusername · Answer

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?

Bunuel · Answer

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.

ashutoshb225 · Answer

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

Bunuel · Answer

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.

sumi747 · Answer

(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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CEdward · Answer

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

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

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