Question

Our school’s girls volleyball team has 14 players, including a set of
3 triplets: Alicia, Amanda, and Anna. In how many ways can we
choose 6 starters if at most one of the triplets is in the starting lineup? There can't be 2 or more triplets and there can be none.

Answer 1

Answer:

$1,\!848$.

Step-by-step explanation:

There are two disjoint sets of ways to choose a lineup as required:

• Include none of Alicia, Amanda, or Anna, or
• Include exactly one of Alicia, Amanda, and Anna.

Assume that none of Alicia, Amanda, or Anna is to be selected. This lineup of $6$ would then need to be selected from a set of $14 - 3 = 11$ players (which excludes Alicia, Amanda, and Anna.)

The number of ways of selecting (without order) $6$ items out of a set of $11$ (distinct) items is equal to the combination:

$\begin{aligned}\begin{pmatrix}11 \\ 6\end{pmatrix} &= \frac{11!}{(6!)\, (11 - 6)!} \\ &= \frac{11!}{6! \times 5!}\end{aligned}$.

Assume that Alicia is selected, but neither Amanda nor Anna is selected. The other $6 - 1 = 5$ players in this lineup would then need to be selected from a set of $14 - 1 - 2 = 11$ players. (This set of $11$ excludes Alicia, Amanda, and Anna.)

The number of ways to select $5$ items from a set of $11$ items is:

$\begin{aligned}\begin{pmatrix}11 \\ 5\end{pmatrix} &= \frac{11!}{(5!)\, (11 - 5)!} \\ &= \frac{11!}{5! \times 6!} \\ &= \frac{11!}{6! \times 5!}\end{aligned}$.

Similarly, there would be another set of $(11!) / (6! \times 5!)$ distinct ways to select the lineup if Amanda is selected, but neither Alicia nor Anna is.

Likewise, the number of ways to select the lineup with Anna but neither Amanda nor Alicia would also be $(11!) / (6! \times 5!)$.

These sets of configurations for the lineup are pairwise disjoint from one another. Thus, the total number of ways to select this lineup would be:

$\begin{aligned}& \begin{pmatrix}11 \\ 6 \end{pmatrix} + 3 \times \begin{pmatrix}11 \\ 5 \end{pmatrix} \\ =\; & \frac{11!}{6! \times 5!} + 3 \times \frac{11!}{6! \times 5!} \\ =\; & \frac{4 \times 11!}{6! \times 5!} \\ =\; & \frac{4 \times 11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \\ =\; & \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 3 \times 2} \\ =\; & 1,\!848\end{aligned}$.