MATHCOUNTS® Problem of the Week Archive Summer Vacation – July 8, 2013 Problems & Solutions Originally posted 7/11/11 John and Olivia Walton and Mike and Carol Brady have plans for their two families to vacation together for one month during the summer. The Waltons will be bringing six of their children. There are three boys, Jason, Ben and Jim‐Bob, and three girls, Mary Ellen, Erin and Elizabeth. The Brady’s will be bringing all six of their children. They also have three boys, Greg, Peter and Bobby, and three girls, Marcia, Jan and Cindy. The two families plan to rent an eight bedroom vacation home on Martha’s Vineyard. John and Olivia will sleep in one of the bedrooms and Mike and Carol will stay in another bedroom. Each of the other six bedrooms will be occupied by a pair of children. The six boys will stay in three of the bedrooms and the six girls will stay in the remaining three bedrooms. Mike Brady has written the numbers 1, 2, 3 and the letters A, B, C on six slips of paper. The papers are folded and placed in a bowl. To randomly determine who will room together, each of the boys will select one of the folded pieces of paper from the bowl. The boys will be paired 1 with A, 2 with B and 3 with C. How many distinct sets of boys roommate pairings are there? Each set of roommate pairings consists of 3 pairs of boys. There are 6C2 = 15 ways of selecting the first
roommate pair. There are 4C2 = 6 ways of selecting the second roommate pair in the set. That leaves just
2C2 = 1 way of choosing the last roommate pair of the set. That means there are 15 × 6 × 1 = 90 sets of
roommate pairings. But they are not all distinct. Consider the following 6 sets which are among the group
of 90 sets:
Jim-Bob & Jason
Ben & Greg
Peter & Bobby
Peter & Bobby
Jim-Bob & Jason
Ben & Greg
Ben & Greg
Peter & Bobby
Jim-Bob & Jason
Jim-Bob & Jason
Peter & Bobby
Ben & Greg
Ben & Greg
Jim-Bob & Jason
Peter & Bobby
Peter & Bobby
Ben & Greg
Jim-Bob & Jason
As you can see these six sets of pairings are not distinct. Each distinct set of pairings can actually occur in
3! = 6 different ways. So there are 90  6 = 15 distinct pairings.
What is the probability that Jim‐Bob will not room with one of his brothers? Express your answer as a percent. Jim-Bob is equally likely to room with any of the other five boys. Three of these boys are not his brothers.
Sothe probability that Jim-Bob will not room with one of his brothers is 3/5 = 60%.
After the boys have randomly selected a roommate, the process is repeated to determine which girls will room together. The girls’ ages are listed in the table below: Marcia 13 Mary Ellen 13 Erin 12 Jan 10 Cindy 8 Elizabeth 6 In what fraction of the total number of distinct sets of the girls roommate pairings is there a roommate pair where the sum of their ages is 23? Express your answer as a common fraction. From the table we see that the sum of the ages for the pairing of Marcia and Jan and the pairing of
Mary Ellen and Jan result in a sum of 23. As with the boys there are a total of 15 distinct sets of roommate
pairings for the girls. If a set of roommate pairings includes the pair Marcia and Jan, the other two pairs can
be selected in 4C2 = 6 ways, but each of these sets can occur in 2 ways, so there are 3 such sets of
roommate pairings where Marcia and Jan are roommates. Similarly, there are 3 distinct sets of roommate
pairings with Mary-Ellen and Jan as roommates. That’s a total of 3 + 3 = 6 distinct sets of roommate pairings.
Therefore, 6/15 = 2/5 of the distinct sets of girls pairings result in a pair where the sum of their ages is 23.