Question

Thirteen people on a softball team show up for a game.

a) How many ways are there to choose \({\bf{1}}0\) players to take the field?

b) How many ways are there to assign the \({\bf{1}}0\) positions by selecting players from the \({\bf{1}}3\) people who show up?

c) Of the\({\bf{1}}3\) people who show up, three are women. How many ways are there to choose \({\bf{1}}0\) players to take the field if at least one of these players must be a woman?

Short answer

a) The possible number of ways is \(286\).

b) The possible number of ways is \(1,037,836,800\).

c) The possible number of ways is \(285\).

Step-by-step solution

Given data

Number of people \( = 13\) , number of players take the field \( = 10\) and number of women \( = 3\).

Concept of Combination

A combination is a selection of items from a set that has distinct members.

Formula:

\(_n{C_r} = \frac{{n!}}{{r!(n - r)!}}\)

Calculation to choose \({\bf{1}}0\) players to take the field

a)

Number of people \( = 13\)

Number of players take the field \( = 10\)

Find the number of ways to choose \(10\) players, use the formula of combination:

Number of ways \( = C(13,10)\)

\(\begin{array}{l}C(13,10) = \frac{{13!}}{{10!(13 - 10)!}}\\C(13,10) = \frac{{13!}}{{10!3!}}\\C(13,10) = 286\end{array}\)

Hence, the possible number of ways is \(286\).

Calculation to assign the \({\bf{10}}\) positions by selecting players from the \({\bf{1}}3\) people

b)

Number of people \( = 13\)

Number of positions \( = 10\)

Find the number of ways to assign \(10\) positions, use the formula of permutation:

Number of ways \( = P(13,10)\)

\(\begin{array}{l}P(13,10) = \frac{{13!}}{{(13 - 10)!}}\\P(13,10) = \frac{{13!}}{{3!}}\\P(13,10) = 1,037,836,800\end{array}\)

Hence, the possible number of ways is \(1,037,836,800\).

Calculation to choose \({\bf{10}}\) players and at least one player is woman

c)

Find the number of ways in which one of these players must be woman, use the formula of combination:

Number of ways \( = 1\) woman and \(9\) men \( + 2\) women and \(8\) men \( + 3\) women and \(7\) men

\(\begin{array}{l}C(10,9) \times C(3,1) + C(10,8) \times C(3,2) + C(10,7) \times C(3,3) = \frac{{10!}}{{9!(10 - 9)!}} \times \frac{{3!}}{{1!(3 - 1)!}} + \frac{{10!}}{{8!(10 - 8)!}} \times \frac{{3!}}{{2!(3 - 2)!}} + \frac{{10!}}{{7!(10 - 7)!}} \times \frac{{3!}}{{3!(3 - 3)!}}\\\;C(10,9) \times C(3,1) + C(10,8) \times C(3,2) + C(10,7) \times C(3,3) = 10 \times 3 + 45 \times 3 + 120 \times 1\\\;C(10,9) \times C(3,1) + C(10,8) \times C(3,2) + C(10,7) \times C(3,3) = 285\end{array}\)

Hence, the possible number of ways is \(285\).