can the following sum be simplified For $n \ge 3$, define
$$f(n) = \sum_{k=3}^n {n \choose k}{k-1 \choose 2}.$$
Is there a closed form expression for $f$?
 A: Note that
$$\binom{n}k\binom{k-1}2=\frac12\binom{n}k(k-1)(k-2)\;,$$
where the $(k-1)(k-2)$ looks like the coefficient of the second derivative of $x^{k-1}$. That suggests looking at something like
$$g(x)=\sum_{k=3}^n\binom{n}kx^{k-1}$$
and differentiating twice with respect to $x$ to get
$$g''(x)=\sum_{k=3}^n\binom{n}k(k-1)(k-2)x^{k-2}\;:$$
clearly we then have $f(n)=\frac12g''(1)$, and all that remains is to get a closed form for $g(x)$. But $g(x)$ can be written
$$g(x)=\frac1x\sum_{k=3}^n\binom{n}kx^k\;,$$
and you know a closed form for $\sum_{k=0}^n\binom{n}kx^k$, so all that’s needed now is a little algebra.
A: We will use the identities
$$
\begin{align}
\sum_{k=m}^n\binom{n}{k}\binom{k}{m}
&=\sum_{k=m}^n\binom{n}{m}\binom{n-m}{k-m}\\
&=\binom{n}{m}2^{n-m}\tag{1}
\end{align}
$$
and
$$
\binom{k-1}{2}=\binom{k}{2}-\binom{k}{1}+\binom{k}{0}\tag{2}
$$
noting that
$$
\binom{k-1}{2}=\left\{\begin{array}{}
0&\text{for }k\in\{1,2\}\\
1&\text{for }k=0
\end{array}\right.\tag{3}
$$
Applying $(1)$, $(2)$, and $(3)$, we get
$$
\begin{align}
\sum_{k=3}^n\binom{n}{k}\binom{k-1}{2}
&=-\binom{n}{0}\binom{-1}{2}+\sum_{k=0}^n\binom{n}{k}\binom{k-1}{2}\\
&=-1+\sum_{k=0}^n\binom{n}{k}\left(\binom{k}{2}-\binom{k}{1}+\binom{k}{0}\right)\\
&=-1+\binom{n}{2}2^{n-2}-\binom{n}{1}2^{n-1}+\binom{n}{0}2^{n-0}\\[9pt]
&=(n^2-5n+8)2^{n-3}-1\tag{4}
\end{align}
$$
A: Wolfram|Alpha says:$$\sum_{k=3}^n\binom{n}{k}\binom{k-1}{2}=2^{n-3}n^2-5\cdot2^{n-3}n+2^n-1$$
A: Use the Binomial Theorem to write down the expansion of
$$\frac{(1+x)^n-1}{x}.$$
Then differentiate twice, and set $x=1$. You will get a very close relative of your sum.  
