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August 8th, 2008, 10:29 PM  #1 
Newbie Joined: Aug 2008 Posts: 1 Thanks: 0  Alternative Permutation proof
This is an alternative proof of permutation but I am having a hard time fully understanding it: We may enumerate , exhaustively and without repetition, the rpermutation as follows: *1st. All those in which the first letter stands first; 2nd. All those in which stands first: and so on. *There are as many permutations in which stands first as there are (r1)permutation of the remaining letters, that is, there are permutations in the first class. The same is true of each of the other classes. *Hence = Hence we may write successively: = , , ……………………………………………. If now we multiply all these equations together and observe that all P’s cancel each other except and , and observe further that the value of is obviously , we see that I can follow the last part of the proof but am having a hard time understanding the sections with *. 1) What does “rpermutation” really mean anyway? 2) Why is that “There are as many permutations in which stands first as there are (r1)permutation of the remaining letters”? Any comments at all will be appreciated! 
August 9th, 2008, 01:00 AM  #2 
Member Joined: Jul 2008 From: Minnesota, USA Posts: 52 Thanks: 0  Re: Alternative Permutation proof
1) By rpermutation they mean choosing (without replacement) r numbers from, in this case, the first n numbers. Order matters. 2) Look at all such permutations starting with 1. Then you still have to choose r1 from n1 other numbers. This essentially reduces the problem down by 1 in both n and r. The motivation here is that we know what nP1 is, it is just n (or you could even go one more step and look at nP0 which is 1). I agree that the wording is a little confusing, but I don't know if I simplified it at all. Nathan 

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