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August 8th, 2008, 10:29 PM   #1
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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 r-permutation 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 (r-1)-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 “r-permutation” really mean anyway?
2) Why is that “There are as many permutations in which stands first as there are (r-1)-permutation of the remaining letters”?

Any comments at all will be appreciated!
kramps77 is offline  
August 9th, 2008, 01:00 AM   #2
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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.

NClement is offline  

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