Question 74205: I have the following problem:
Let P(n, k) denote the number of permutations of k objects selected from a set of n. We have the formula P (n,k) = n!/(n-k)!
Prove that for all integers n>or=2, P(n+1,2)-P(n, 2)=2P(n, 1).
I did some research on factorial identities and got a little further, but I'm not getting to the desired end result. I guess where I'm most stuck is the goal -- is 2P(n,1) equal to 2(n!/(n-1)!)? I've never seen the '2P' before. Thanks!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I guess where I'm most stuck is the goal -- is 2P(n,1) equal to 2(n!/(n-1)!)?
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Yes, since P(n,1) = n!/(n-1)!, 2P(n,1) is 2[n!/(n-1)!]
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After showing the form works for n=2, and assuming it is true for n=k,
you want to show that if n=k+1, P(k+2,2)-P(k+1,2) = 2P(k+1,1)
That can be written as:
[(k+2)!/(k)!] - [(k+1)!/(k-1)!] = 2[(k+1)!/k!]
(k+2)(k+1) - (k+1)k =2[(k+1)]
k^2+3k+2 -k^2-k = 2k+2
2k+2 = 2k+2
That's the proof.
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Cheers,
Stan H.
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