Binomial probability with summation Show that
$$\sum_{k=0}^{m} \frac{m!(n-k)!}{n!(m-k)!} = \frac{n+1}{n-m+1}$$
Attempt:
It becomes:
$$\sum_{k=0}^{m } \frac{\binom{m}{k}}{\binom{n}{k}}$$
Telescoping, pairing, binomial theorem don't seem to work
Possibly a combinatoric proof?
Considering the cards of numbers 1, 2, 3, 4, ... , m, .... , n on the table. 
We must pick k cards from the n on the table. 
The probability that we pick these k cards from the first m is
$$P(k) = \frac{\binom{m}{k}}{\binom{n}{k}}$$
So summing these probabilities will give us the LHS.
Now is there a nice way of doing this probability a different way to yield the RHS?

If I write your sum as a hypergeometric series, it becomes $$\sum_{k=0}^{m} \frac {m! \left(n-k\right)!}{n! \left( m-k\right)!}={}_2F_1\left( \left.\begin{array}{c} -m, 1\\-n \end{array} \right| 1 \right)$$
If $m \le n$, this is a special case of the Chu-Vandermonde identity, which will give you the desired result. You may also google "Chu-Vandermonde identity" for a combinatorial proof. It may give you a clue for a combinatorial proof of your identity.
