how can i prove the equality in hypergeometric distribution?

Question

the equality is:
sigma(from j=0 to N) C(N,j)*C(M,n-j)=C(N+M,n)
I used the identity of (1+t)^(N+M)=(1+t)^N+(1+t)^M
but I wasn't able to solve it
thanks...

edit for identity: sorry my identity would be like that
(1+t)^(M+N)=((1+t)^(M))*((1+t)^(N))
still I can't get the equality.Could you clarify it clearly?

Answer 1

Write out explicitly what nCk is in terms of factorials and use communtitive property to move factors around

NCj = N!/[(N-j)!j!]

do the same for MC(n-j)

multiply them and move factors around. You might need to introduce another factor to use the symmetry property:
nCk=nC(n-k) (you can see this clearly on Pascal's triangle, or just write it out in terms of factorials)

+Mon
I found this hint in Marx & Larsen:

expand
(1+u)^N = [(1+u)^r]*[(1+u)^(N-r)]
and equated coefficients.
(that sorta makes sense) Note that if you had
(a+b)^(N+M) =[ (a+b)^N]*[(a+b)^M]
you'd equated coeff of (a^j)(b^k)

Answer 2

If a=b then you may not divide by making use of a-b in step 3 because of the undeniable fact that is dividing by making use of 0 that's a no-no. Is it basically me? Or do those comparable varieties of uninteresting, dumbshit questions on dividing by making use of 0 and what's 0/0 start to get slightly tedious after seeing them 2 or thrice in keeping with day? Doug

Answer 3

If your C(N,j) is the binomial coefficient, the correct sum is
sigma(from j=0 to n) C(r,j)*C(s,n-j)=C(r+s,n) for r+s>=n
which differs from yours. I believe that there is a proof by recurrsion.