Explain Why ${21 \choose 2}^2 - {21 \choose 2} = 3!{22 \choose 4}$ I was given this little problem for precalc homework after a class discussion on series and sigma notation, and applying combinatorial approaches to them. We happened upon the equation in a larger problem we were doing, and the teacher was surprised that it was true and suspected that there might be an underlying combinatorial explanation for it, which he challenged us to find. I suspect it might have something to do with the Hockey Stick Theorem, which we had explored earlier in class.
 A: A more combinatorial view:
Suppose you had 21 people. You can then form $21\choose 2$ unordered pairs of people, which for the rest of the answer I will call a pairing. Suppose you wanted to count the number of ordered pairs of pairings such that the two pairings are not the same (they may share up to one person, but no more).
One way to do this is to note that there are $21\choose 2$ possible pairings, so the number of ordered pairs of different pairings is $\left({21\choose 2}\right)\left({21\choose 2} -1\right)$. This is the left-hand side of your equation.
A different way is the following: introduce an additional person X who will not be part of the pairings in the end, and simply choose 4 people from the enlarged group of 22. Now form the ordered pair of pairings as follows:


*

*If person X is not among the 4 chosen, then simply choose 2 to be in the first pairing, leaving the other 2 in the second pairing. There are ${4\choose 2} = 6$ ways to do this.

*If person X is among the 4 chosen, then choose one of the 3 others to be in both pairings, and randomize the order of the other 2. There are $3\times 2 = 6$ ways to do this.


It is fairly easy to see that this selection process yields all ordered pairs of pairings, since there are always at least 3 distinct people among any ordered pair of pairings. The number of ways to perform this selection process is ${22\choose 4}\times 6$, which is your right-hand side.
Replacing $21$ with $n$ and $22$ with $n+1$ yields a more general result.
A: The general identity behind your result:
$$  \forall n \in \mathbb{N} \ \ \dfrac{n(n - 1)}{2}\left(\dfrac{n(n - 1)}{2} - 1\right) = \dfrac{(n + 1)n( n - 1)(n - 2)}{4}$$
A: Here is a somewhat awkward combinatorial interpretation that works for general $n\ge 3$. Players A and B each choose an unordered pair of numbers from the numbers $1$ to $n$. How many ways can it happen that Players A and B chose different pairs? 
It is clear that the number of ways is $\binom{n}{2}^2 -\binom{n}{2}$. 
We now count another way. Add to our collection $\{1,2,\dots,n\}$ the abstract object $\ast$, and choose $4$ objects from this extended set. This can be done in $\binom{n+1}{4}$ ways.
If we do not choose $\ast$ among the $4$, the $4$ objects can be distributed between A and B in $3!$ ways. If $\ast$ is chosen, that is an indication  that one of the other three numbers chosen is to be in both A's hand and B's hand. In this case again, the distribution can be done in $3!$ ways. This gives a count of $3!\binom{n+1}{4}$ ways for A and B to choose two different pairs of numbers from $\{1,2,\dots,n\}$.
A: $$\binom {n}{2}=n(n-1)/2 =f(n).$$ $$\binom {n+1}{4}=(n+1)n(n-1)(n-2)/24=g(n).$$ $$3!=6.$$ For all n we have $$f(n)^2-f(n)=(n^4-2 n^3-n^2+2 n)/4=6 g(n).$$
A: We have 
$\binom{21}{2}=210$  and $\binom{22}{4}=7315$. Then 
$$
\binom{21}{2}^2-\binom{21}{2}=210^2-210=210\cdot (210-1)=43890=6 \cdot 7315=3! \binom{22}{4}.
$$
