using generating functions to prove vandermonde's identityC (m +n, r) = ∑r k=0 C(m,r- k) C(n,k) whenever m, n and r are nonnegative integers with r not exceeding either m or n

Using generating functions, Vandermonde's identity can be proven as C(m+n,r) = ∑r k=0 C(m,r-k) C(n,k), where C(n,k) denotes the binomial coefficient. This identity is useful in combinatorics and probability theory, as it provides a way to calculate the number of combinations of r objects that can be chosen from two sets of m and n objects.

To use generating functions to prove Vandermonde's identity, we can start by defining two generating functions:

f(x) = (1+x)^m
g(x) = (1+x)^n

Using the binomial theorem, we can expand these generating functions as:

f(x) = C(m,0) + C(m,1)x + C(m,2)x^2 + ... + C(m,m)x^m
g(x) = C(n,0) + C(n,1)x + C(n,2)x^2 + ... + C(n,n)x^n

Now, let's multiply these two generating functions together and look at the coefficient of x^r:

f(x)g(x) = (1+x)^m (1+x)^n = (1+x)^(m+n)

Expanding this using the binomial theorem gives:

f(x)g(x) = C(m+n,0) + C(m+n,1)x + C(m+n,2)x^2 + ... + C(m+n,m+n)x^(m+n)

So, the coefficient of x^r in f(x)g(x) is equal to C(m+n,r).

Now, let's rearrange the terms in f(x)g(x) to isolate the term involving C(m,r-k) and C(n,k):

f(x)g(x) = (C(m,0)C(n,r) + C(m,1)C(n,r-1) + ... + C(m,r)C(n,0))x^r
         + (C(m,0)C(n,r+1) + C(m,1)C(n,r) + ... + C(m,r+1)C(n,0))x^(r+1)
         + ...

So, the coefficient of x^r in f(x)g(x) is also equal to the sum:

∑r k=0 C(m,r- k) C(n,k)

Therefore, we have shown that C(m+n,r) = ∑r k=0 C(m,r- k) C(n,k), which is Vandermonde's identity.

Learn more about Vandermonde's identity:

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