## Round 1: Problem 2 Solution

This problem at first glance appears somewhat cumbersome, but with an easy trick the answer should be found very quickly.

In contrast to the way the problem is written, for simplicity’s sake let us ignore Team B and simply focus on Team A. When Team A wins, we will denote the game with a W. When A loses, an L. Then, we should end up with some combination of W’s and L’s, such as WLWLLWW. Each of these strings of W’s and L’s correspond to a sequence of wins and losses that may occur in the series.

The Principle of Inclusion / Exclusion really should not be needed here. It works, but it’s not necessary. The easiest way to solve the problem is to think about how problematic combinations occur only when a L occurs in the final game. Note that the series can extend to 4, 5, 6, or 7 games. Then to eliminate the problematic combinations, let us force ourselves to pick a W in the final spot. Now all we have to do is consider each series length and pick the other three games that Team A is going to win.

For 4 games, besides the W that was already chosen in the 4th game, there are 3 remaining games, in which we need to place 3 more W’s, which can be done in **3C3 = 1 way.**

For 5 games, 4 games remain to place the 3 W’s in **4C3 = 4 ways.**

For 6 games, 5 remaining games for the 3 W’s in **5C3 = 10 ways.**

Finally, for 7 games, 6 remaining games for the 3 W’s in **6C3 = 20 ways.**

In total, that sums to 35 ways. We now multiply by 2 to compensate for the fact that either Team A or Team B can win in any of these ways, giving the final answer of **70 ways** for the sequence to occur.

As always, please post any comments, corrections or alternate solutions in the comments. In particular if you’ve worked out the Inclusion / Exclusion, might as well post it.

Actually, we can make it even simpler than that: every sequence of games with 4 wins and 3 or fewer losses can be extended to a sequence of games with 4 wins and exactly 3 losses in a unique way. For example, WWLWW extends to WWLWWLL, WLLWWW extends to WLLWWW and WWLLLWW extends to itself. Thus, the answer is 2 * 7C4. The equality 3C3 + 4C3 + 5C3 + 6C3 = 7C4 is one instance of the so-called “hockey-stick identity.”

JBLMarch 17, 2009 at 17:45

I always forget about that identity… Thank you!

AdministratorMarch 17, 2009 at 23:06

Great post, but its a bit long and most people like short and sweet posts!

SimonnMarch 21, 2009 at 23:48