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Permutations

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Permutations

by apex231 » Thu Aug 18, 2011 7:49 pm
In how many ways the word 'PERMUTATIONS' can be arranged such that 4 alphabets exist between the alphabet P and S?

I don't have an answer to this question.

Following is the approach that I followed. Is it correct or any other better approach to solve this problem?

Number of ways P and S can be placed such that 4 alphabets are present between them = 8 i.e. (1,5) (2,6) (3,7) (4,8) (5,9) (6,10) (7,11) (8,12)

Similarly, number of ways S and P can be placed such that 4 alphabets are present between them = 8

Total number of ways = 8 * 2 = 16

Remaining 10 positions can be occupied in 10 P(10) = 10!

Total permutations = 16 * 10!
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by Frankenstein » Thu Aug 18, 2011 8:05 pm
Hi,
I guess there are a couple of minor mistakes.
P,_,_,_,_,S
1,2,3,4,5,6
If there are 4 alphabets between P and S, their positions should be (1,6),(2,7)....(7,12) -> 7 ways
As positions of S and P are interchangeable, we get 2*7 = 14 ways
Now, in the remaining 10 alphabets, there are two T's. So, they can be arranged in 10!/2! ways
So, I guess the answer should be 14*10!/2! = 7*10!.
Btw, there are other approaches, but this is just fine as it is simple and easy to understand.
Cheers!

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