Is it true that $\sum_{k=3}^{n}\binom{n}{k}\binom{k}{3} = \binom{n}{3}\binom{n-3}{k-3}$? I was asked to find a closed formula for $$\sum_{k=3}^{n}\binom{n}{k}\binom{k}{3}$$
To remove the $\sum$ if you will.
Here's my reasoning, let's say we have a football team with $n$ players. First we choose $k$ players from those $n$ to be on the starting lineup, and then we chose $3$ out of those $k$ players to play defense.
That is what $\sum_{k=3}^{n}\binom{n}{k}\binom{k}{3}$ means (I think).
Now, that is the same thing as first choosing $3$ out of the initial $n$ to play defense, and then choosing $k-3$ out of the remaining $n-3$ to be on the starting lineup, and that's $\binom{n}{3}\binom{n-3}{k-3}$, so I inferred from that:
$$\sum_{k=3}^{n}\binom{n}{k}\binom{k}{3} = \binom{n}{3}\binom{n-3}{k-3}$$ is this true?
 A: Your football team combinatorial argument, with some modification, will work. We have a team of $n$ people. We want to select $3$ of them to get a gold medal, and a subset (possibly empty) of the rest to get a silver medal. 
We count the number of ways to do this in two different ways.
(1) The gold medal winners can be chosen in $\binom{n}{3}$ ways. For every such way, the set of silver medal winners can be chosen in $2^{n-3}$ ways, for a total of $\binom{n}{3}2^{n-3}$ ways.
(2) For any $k\ge 3$, we can choose the $k$ people who win some medal in $\binom{n}{k}$ ways, and for every choice we can choose the gold medal winners from these in $\binom{k}{3}$ ways, for a total of $\binom{n}{k}\binom{k}{3}$ ways. Since we are free to choose any number $k\ge 3$ to get some medal, the total number of ways to select the gold and silver medal winners is $\sum_{k=3}^n \binom{n}{k}\binom{k}{3}$. 
A: You've proved a version of the trinomial revision identity.  However, you accidentally dropped the summation; what is true is  $${n\choose k}{k\choose 3}={n\choose 3}{n-3\choose k-3}$$
Now you may pull the ${n\choose 3}$ out of the sum, reindex, and use another well-known identity on binomial coefficients.
