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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 $_{n} P_{r}$ r-permutation as follows: *1st. All those in which the first letter $a_{1}$ stands first; 2nd. All those in which $a_{2}$ stands first: and so on. *There are as many permutations in which $a_{1}$ stands first as there are (r-1)-permutation of the remaining $n-1$ letters, that is, there are $_{n-1}P_{r-1}$ permutations in the first class. The same is true of each of the other $n$classes. *Hence $_{n} P_{r}$ = $_{n-1}P_{r-1}$ Hence we may write successively: $_{n} P_{r}$ = $_{n-1}P_{r-1}$, $_{n-1} P_{r-1}= (n-1) _{n-2} P_{r-2}$, ……………………………………………. $_{n-r+2} P_{2}= (n-r+2) _{n-r+1} P_{1}$ If now we multiply all these equations together and observe that all P’s cancel each other except $_{n} P_{r}$ and $_{n-r+1} P_{1}$ , and observe further that the value of $_{n-r+1} P_{1}$ is obviously $n-r+1$, we see that $_{n} P_{r}= n(n-1)(n-2) ... (n-r+2)(n-r+1)$ I can follow the last part of the proof but am having a hard time understanding the sections with *. 1) What does “r-permutation” really mean anyway? 2) Why is that “There are as many permutations in which $a_{1}$ stands first as there are (r-1)-permutation of the remaining $n-1$ 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 r-permutation 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 r-1 from n-1 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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