I'm in an introductory discrete mathematics course and i'm having a lot of trouble with this proof. I'm not sure if i'm heading in the right direction or not but I used the fact that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ and started trying to algebraically reduce them to eachother but haven't succeeded. Any help would be much appreciated.
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The algebraic reduction should be straightforward - i.e. just write everything in terms of factorials and both sides should equal $\frac{w!}{(w-p)!m!(p-m)!}$ – Michael Biro Mar 23 '19 at 17:55
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Wow you are right I accidentally wrote $\binom{p}{m} = \frac{p!}{m!(p-w)!}$ and never checked that part (at the beginning) thanks so much. – Tiop Mar 23 '19 at 18:01