usre123 wrote:
Bunuel wrote:
usre123 wrote:
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
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 = \(\frac{6!}{2!}*C^5_7*\frac{5!}{3!}\).
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?
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
ION, 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.
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