A Combinatorial Proof for (n choose 2) choose 2 = 3(n choose 4) + 3(n choose 3)

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SUMMARY

The discussion focuses on proving the combinatorial identity \(\binom{n}{2}\) choose 2 = 3\(\binom{n}{4}\) + 3\(\binom{n}{3}\). The left-hand side represents the number of ways to select two unordered pairs from a total of \(\binom{n}{2}\) pairs, which can be interpreted as counting unordered pairs of unordered pairs. The right-hand side breaks down the selection into cases involving either four distinct elements or three distinct elements, leading to the conclusion that the identity holds true through combinatorial reasoning.

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clooneyisagen
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Homework Statement



((n choose 2) choose 2) = 3(n choose 4) + 3(n choose 3)

Need a combinatorial proof...

Homework Equations


for example, (n choose k) means from a total of n people we choose a committe of size k.
(though this may not be relevant equation)


The Attempt at a Solution


I'm thinking for the left hand side that out of a total of n people we find the all the possible committees of size 2. Then of all these committees we pick two of the duos picked? No idea how to do the right hand side - or make the left side equal
 
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The crucial point, then, is "what does
\left(\begin{array}{c}\left(\begin{array}{c} n \\ 2 \end{array}\right) \\ 2 \end{array}\right)
mean in terms of combinatorics"?
It would be, apparently, the number of different ways to choose 2 people from a group of \left(\begin{array}{c} n \\ 2\end{array}\right)
Now, how would you interpret that \left(\begin{array}{c} n \\ 2\end{array}\right) "combinatorically"?
 
Well the left hand side \binom{\binom{n}{2}}{2} counts the number of unordered pairs of unordered pairs. What exactly can one look like? Either it involves 4 distinct elements, or it involves 3 distinct elements, like {{a,b}, {a,c}}, but it can never involve 2 or 1 distinct elements (why)? I hope this is enough to push you in the right direction.
 

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