factorial question
Author
Can someone show me how to simplify this factorial:
(n-1)! / ( (n-i-1)! i! )
which results in to:
1/(n-i) + 1/i
1. Is this homework? Forum FAQ
2. Are you sure it reduces to that? I'm not going to work it but that seems to be a mighty peculiar answer.
2. Are you sure it reduces to that? I'm not going to work it but that seems to be a mighty peculiar answer.
Author
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.
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?
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?
This topic is closed to new replies.
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