Subfield: Combinatorics


Question: Eight people, including triplets Barry, Carrie and Mary, are going for a trip in four canoes. Each canoe seats two people. The eight people are to be randomly assigned to the four canoes in pairs. What is the probability that no two of Barry, Carrie and Mary will be in the same canoe?

Solution:
Among a group of $n$ people, there are $\frac{n(n-1)}{2}$ ways of choosing a pair of these people:

There are $n$ people that can be chosen first.

For each of these $n$ people, there are $n-1$ people that can be chosen second.

This gives $n(n-1)$ orderings of two people.

Each pair is counted twice (given two people A and B, we have counted both the

pair $\mathrm{AB}$ and the pair $\mathrm{BA})$, so the total number of pairs is $\frac{n(n-1)}{2}$.

We label the four canoes W, X, Y, and Z.

First, we determine the total number of ways to put the 8 people in the 4 canoes.

We choose 2 people to put in W. There are $\frac{8 \cdot 7}{2}$ pairs. This leaves 6 people for the remaining 3 canoes.

Next, we choose 2 people to put in X. There are $\frac{6 \cdot 5}{2}$ pairs. This leaves 4 people for the remaining 2 canoes.

Next, we choose 2 people to put in Y. There are $\frac{4 \cdot 3}{2}$ pairs. This leaves 2 people for the remaining canoe.

There is now 1 way to put the remaining people in $\mathrm{Z}$.

Therefore, there are

$$
\frac{8 \cdot 7}{2} \cdot \frac{6 \cdot 5}{2} \cdot \frac{4 \cdot 3}{2}=\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{2^{3}}=7 \cdot 6 \cdot 5 \cdot 4 \cdot 3
$$

ways to put the 8 people in the 4 canoes.

Now, we determine the number of ways in which no two of Barry, Carrie and Mary will be in the same canoe.

There are 4 possible canoes in which Barry can go.

There are then 3 possible canoes in which Carrie can go, because she cannot go in the same canoe as Barry.

There are then 2 possible canoes in which Mary can go, because she cannot go in the same canoe as Barry or Carrie.

This leaves 5 people left to put in the canoes.

There are 5 choices of the person that can go with Barry, and then 4 choices of the person that can go with Carrie, and then 3 choices of the person that can go with Mary.

The remaining 2 people are put in the remaining empty canoe.

This means that there are $4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 \cdot 3$ ways in which the 8 people can be put in 4 canoes so that no two of Barry, Carrie and Mary are in the same canoe.

Therefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $\frac{4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 \cdot 3}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}=\frac{4 \cdot 3 \cdot 2}{7 \cdot 6}=\frac{24}{42}=\frac{4}{7}$.
Let $p$ be the probability that two of Barry, Carrie and Mary are in the same canoe.

The answer to the original problem will be $1-p$.

Let $q$ be the probability that Barry and Carrie are in the same canoe.

By symmetry, the probability that Barry and Mary are in the same canoe also equals $q$ as does the probability that Carrie and Mary are in the same canoe.

This means that $p=3 q$.

So we calculate $q$.

To do this, we put Barry in a canoe. Since there are 7 possible people who can go in the canoe with him, then the probability that Carrie is in the canoe with him equals $\frac{1}{7}$. The other 6 people can be put in the canoes in any way.

This means that the probability that Barry and Carrie are in the same canoe is $q=\frac{1}{7}$.

Therefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $1-3 \cdot \frac{1}{7}$ or $\frac{4}{7}$.

Final answer: $\frac{4}{7}$