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?
 A: I computed this sum for this answer by induction on $m$:
The formula is trivially true for $m=0$. Assume it is true for $m-1\le n-1$:
$$
\begin{align}
\sum_{k=0}^m\frac{\binom{\vphantom{1+}m}{k}}{\binom{\vphantom{1+}n}{k}}
&=1+\sum_{k=1}^m\frac{\frac{\vphantom{1+}m}{k}\binom{m-1}{k-1}}{\frac{n}{k}\binom{n-1}{k-1}}\\
&=1+\frac{m}{n}\sum_{k=0}^{m-1}\frac{\binom{m-1}{k}}{\binom{n-1}{k}}\\[6pt]
&=1+\frac{m}{n}\frac{n}{n-m+1}\\[9pt]
&=\frac{n+1}{n-m+1}
\end{align}
$$
Thus, it is true for all $m\le n$.

However, if we work from Mark Bennet's comment, we get
$$
\begin{align}
\sum_{k=0}^m\frac{\binom{\vphantom{1+}m}{k}}{\binom{\vphantom{1+}n}{k}}
&=\frac1{\binom{\vphantom{1+}n}{m}}\sum_{k=0}^m\binom{n-k}{n-m}\binom{k}{0}\\[6pt]
&=\frac{\binom{n+1}{n-m+1}}{\binom{\vphantom{1+}n}{m}}\\[6pt]
&=\frac{\frac{n+1}{n-m+1}\binom{\vphantom{1+}n}{n-m}}{\binom{\vphantom{1+}n}{m}}\\[6pt]
&=\frac{n+1}{n-m+1}
\end{align}
$$
A: 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.
