SOLUTION to Problem 3 Extension
We interpret the left-hand side of the identity as follows. Given n different balls, we first choose 2 of the balls and place a large purple dot on them. We then choose another pair of balls (not exactly the same as the first) and again put large purple dots on them. Then the left-hand side counts the number of ways this can be done; i.e. the number of ways of choosing two pairs of balls.
Now toss a new hot pink ball into the mix. Each method of choosing two pairs of the original n balls will lead to a way of choosing four of these n+1 balls as follows. If the two pairs happened not to overlap then four different balls got purple dots, so we have our selection of four balls. Notice that the marked balls are not sufficient to determine which two pairs were chosen! Thus if balls A, B, C, and D got the purple dots, this could have occurred because we chose pairs (A,B) and (C,D), or pairs (A,C) and (B,D), or pairs (A,D) and (B,C). Hence each set of four balls corresponds to three choices of two pairs in this case.
On the other hand, it is possible that only three balls got purple dots (one of them being marked twice) because the two pairs overlapped somewhat. In this case we declare that the corresponding set of four balls consists of the three marked balls along with the hot pink ball. Once again, just knowing the set of four balls (one of them being hot pink) doesn't tell us which two pairs were originally chosen. Thus the foursome of E, F, G, and H (where H=“hot pink”) might have arisen from the pairs (E,F) and (E,G), or the pairs (E,F) and (F,G), or the pairs (E,G) and (F,G). Once again, each set of four balls corresponds to exactly three choices of two pairs.
In summary, we have shown that every method of choosing four balls from among the n+1 balls corresponds to exactly three ways of choosing two pairs of n balls. This completes the argument.