RE: [math-fun] Re: ... Permutation Probabilities
Hi, Don. Thanks for the explanation. I'm probably missing something, but looking just at our distinct integer expressions for the same probability, it seems non-obvious that they are equal: 109*108*...*92*91 126*124*122*120*...* 92 Why would ----------------- = ----------------------- ? 127*125*... 93*91 127*126*125*124*...*110 --Dan
On Wed, 9 Jun 2004, Daniel Asimov wrote:
Hi, Don.
Thanks for the explanation.
I'm probably missing something, but looking just at our distinct integer expressions for the same probability, it seems non-obvious that they are equal:
109*108*...*92*91 126*124*122*120*...* 92 Why would ----------------- = ----------------------- ? 127*125*... 93*91 127*126*125*124*...*110
--Dan
Well, it follows from the well-known (:-)) identity:
product(i,i=2*p+1..p+q+1)/product(2*i+1,i=p..q) = product(2*i,i=p+1..q)/product(i,i=p+q+2..2*q+1);
which Maple reduces to: p / 1\ GAMMA(p + q + 2) 2 GAMMA|p + -| (q + 1) \ 2/ 2 GAMMA(q + 1) GAMMA(p + q + 2) ------------------------------------ = -------------------------------------- (q + 1) / 3\ (p + 1) GAMMA(2 p + 1) 2 GAMMA|q + -| 2 GAMMA(p + 1) GAMMA(2 q + 2) \ 2/ and verifies equality:
simplify(lhs(%)-rhs(%)); 0
--Edwin
participants (2)
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Daniel Asimov -
Edwin Clark