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LordAdkin

factorial question

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No it's not home work. I'm just looking at a solution to an exam (http://www.dgp.toronto.edu/~hertzman/courses/csc418/winter_2003/final-solutions.pdf), and I don't know how that factorial got reduced to that form. I haven't done factorial stuff in quite a while, so I'm just wondering how it's done.

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i'd say start by substituting m as n-1 to give
m! / ( (m-i)! i! )

but I think your answer is incorrect as...

substituting n as 6 and i as 2 gives...

(n-1)! / ( (n-i-1)! i! )
(6-1)! / ( (6-2-1)! 2! )
5! / (3!2!)
120 / 12
10

1/(n-i) + 1/i
1/(6-2) + 1/2
1/(4) + 1/2
3/4

Clearly 10 != 3/4
Are you sure thats the right answer?

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