1. ## Binomial summation proof

Show that:

$\displaystyle \sum_{k=m}^n \binom{k}{m}\binom{n}{k}p^k(1-p)^{n-k}=\binom{n}{m}p^m$

You may be able to use this result:

$\displaystyle \binom{k}{m}\binom{n}{k}=\binom{n}{m}\binom{n-m}{k-m}$

2. Originally Posted by acevipa
Show that:

$\displaystyle \sum_{k=m}^n \binom{k}{m}\binom{n}{k}p^k(1-p)^{n-k}=\binom{n}{m}p^m$

You may be able to use this result:

$\displaystyle \binom{k}{m}\binom{n}{k}=\binom{n}{m}\binom{n-m}{k-m}$

$\displaystyle (x+y)^n = \sum_k \binom{n}{k}x^k y^{n-k}$

Differentiate m times with respect to x:

$\displaystyle (n-m+1) \cdots (n-1)(n) (x+y)^{n-m} = \sum_k \binom{n}{k} (k-m+1) \cdots (k-1)(k) x^{k-m}y^{n-k}$

Divide by m!:

$\displaystyle \binom{n}{m} (x+y)^{n-m} = \sum_k \binom{n}{k} \binom{k}{m} x^{k-m}y^{n-k}$

Now let $\displaystyle x = p$ and $\displaystyle y = 1-p$:

$\displaystyle \binom{n}{m} (1)^{n-m} = \sum_k \binom{n}{k} \binom{k}{m} p^{k-m}(1-p)^{n-k}$

Multiply by $\displaystyle p^m$:

$\displaystyle \binom{n}{m} p^{m} = \sum_k \binom{n}{k} \binom{k}{m} p^k (1-p)^{n-k}$