## Saturday, September 24, 2011

### May Problem and Solution!!!

Hey Guys! In May of 2011 the Bard Math Circle visited the Kingston Library and challenged the enthusiastic students to solve a series of combination/permutation problems. Although tough, the problems proved to be quite the adventure for the talented youth. One of the problems asked to find the different ways that four friends could split into two group (groups can be uneven) if they were to play a game. Here is the problem:

Four friends, Andrew, Brianna, Cassandra, and Darien want to
play a game. How many ways can they form two teams? Every-
one must play.
Example: Andrew, Cassandra, Darien vs. Brianna .
This is the same as Brianna vs. Darien, Andrew, Cassandra

Actually, this is a similar situation to my 2011 Summer at Bard College adventures with three friends of mine (Anam, Andres, Mauricio). We were having a dance competition and we could either partner up in a group or give a solo performance. We first gave everyone the chance to prove themselves the best dancer. This gave us four groups. Next, Anam got a partner. She had three dances with a partner. Then, Andres had the chance to get a partner as well and he danced only twice (he had already danced with Anam). Later, Mauricio had to get a partner and he only danced with me (he had already danced with Andres and Anam). When it was my turn, I realized I had already danced with all of them and I was too tired of dancing. Nonetheless, it was a fun night and full of laughs, combinations, and math.

It turns out the number of groups we could create was 7. We could abbreviate Andrew as A, Brianna as B, Cassandra as C, and Darien as D. Then the example Andrew, Cassandra, Darien vs. Brianna could be written as ACD|B. To make things easy, let's agree to always list Andrew's team first:

1. A|BCD
2. AB|CD
3. AC|BD