Finding the closed form of a multi-binomial summation I am trying to find the closed form of:
$$\sum_{k=3}^{n} C(n,k)C(k,3)$$
The resulting sequence from $n=3$ to $n=7$ is: 
$$1, 8, 40, 160, 560,\dots$$
After canceling out terms and examining it, the only thing I can solve is
$$\frac{n!}{3!}\cdot(n-2\,\, \mbox{number of terms here})$$
but these terms inside the parenthesis don't have a pattern I can figure out. I might be missing something obvious, however. Factorials aren't quite my thing.
Any tips? Thanks.
 A: The solution is $C(n,3) 2^{n-3}$.  Your sum can be expressed as
$$\sum_{k=3}^n \frac{n!}{(n-k)! 3!(k-3)!} = \sum_{k=0}^{n-3} \frac{n!}{(n-k-3)!3!k!} = C(n,3) \sum_{k=0}^{n-3} C(n-3,k) = C(n,3) 2^{n-3}$$.
A: You can consider a class of $n$ students then answer the following question in two different ways:
Quession: In how many ways can we select at least $3$ student and then select $3$ leaders among selected students?
Answer 1: We can choose $k\geqslant3$ student in $n \choose k$ ways and then choose the $3$ leaders among these $k$ student in $k \choose 3$ ways so by using multiplication and then addition principles the desired number is $$\sum\limits_{k = 3}^n {\left( {\begin{array}{*{20}{c}}
  n \\ 
  k 
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  k \\ 
  3 
\end{array}} \right)}.$$
Answer 2: We can choose the $3$ leaders first in $n \choose 3$ ways then choose any other number of students that we need from remaining $(n-3)$ students in $2^{n-3}$ ways so by multiplication principle the desired number is $$2^{n-3} {n\choose 3}.$$
so we can conclude that
$$\sum\limits_{k = 3}^n {\left( {\begin{array}{*{20}{c}}
  n \\ 
  k 
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  k \\ 
  3 
\end{array}} \right)}=2^{n-3}{n \choose 3}.$$
A: Expanding the binomial coefficients into ratios of factorials, it is easy to show that
$$
\binom{n}{k}\binom{k}{3}=\binom{n}{3}\binom{n-3}{k-3}
$$
Furthermore, the binomial theorem says that
$$
\begin{align}
2^{n-3}
&=(1+1)^{n-3}\\
&=\sum_{k=3}^n\binom{n-3}{k-3}1^{k-3}
\end{align}
$$
Bringing these together yields
$$
\begin{align}
\sum_{k=3}^n\binom{n}{k}\binom{k}{3}
&=\sum_{k=3}^n\binom{n}{3}\binom{n-3}{k-3}\\
&=\binom{n}{3}2^{n-3}
\end{align}
$$
