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Prove the following identity combinatorially

$$\left(\begin{array}{c} \left(\begin{array}{c} n \\ 2 \end{array}\right) \\ 2 \end{array}\right) = 3 \left(\begin{array}{c} n \\ 4 \end{array}\right) + 3 \left(\begin{array}{c} n \\ 3 \end{array}\right) $$

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We want to choose $2$ two-element subsets of an $n$ element set. Clearly there are $\binom{\binom{n}{2}}{2}$ ways to do this.

Let us count another way. Our pairs of doubletons are of two types (i) no overlap or (ii) an overlap of $1$ point.

To produce a pair of doubletons of type (i), choose $4$ objects. They can be split into two doubletons in $3$ ways.

To produce a pair of doubletons of type (ii), choose $3$ objects. There are $3$ ways to decide which of them will be in both doubletons.

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    $\begingroup$ +1 It might be a bit easier to read if the answer is phrased in the language of graph theory. $\endgroup$ – user1551 Mar 6 '13 at 9:27

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