Question

# A software development department consists of 6 women and 4 men. (A) How many ways can the department select a chief programmer, a backup programmer, and a programming librarian? (B) How many of the selections in part (A) consist entirely of women? (C) How many ways can the department select a team of 3 programmers to work on a particular project?

Solution

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$\textbf{(A)}$ As order of selection is relevant, therefore selection is a permutation.

Number of ways to select a chief programmer, a backup programmer, and a programming librarian is:

\begin{align*} _{10}P_3&=\dfrac{10!}{(10-3)!}\\ &=\dfrac{10 \cdot 9 \cdot 8 \cdot \cancel{7!}}{\cancel{7!}}\\ &= 10 \cdot 9 \cdot 8\\ &= 10 \cdot 72\\ &=\textcolor{#4257b2}{720} \end{align*}

$\textbf{(B)}$ As order of selection is relevant, therefore selection is a permutation.

Number of ways to select only women as a chief programmer, a backup programmer, and a programming librarian is:

\begin{align*} _{6}P_3&=\dfrac{6!}{(6-3)!}\\ &=\dfrac{6 \cdot 5 \cdot 4 \cdot \cancel{3!}}{\cancel{3!}}\\ &= 6 \cdot 5 \cdot 4\\ &= 6 \cdot 20\\ &=\textcolor{#4257b2}{120} \end{align*}

$\textbf{(C)}$ As order of selection is irrelevant, therefore selection is a combination.

Number of ways to select 3 programmers is:

\begin{align*} _{10}C_3&=\dfrac{10!}{3!(10-3)!}\\ &=\dfrac{10 \cdot 9 \cdot 8 \cdot \cancel{7!}}{3! \cdot \cancel{7!}}\\ &= \dfrac{10 \cdot \cancel{3} \cdot 3 \cdot \cancel{2} \cdot 4}{\cancel{3} \cdot \cancel{2} \cdot 1}\\ &= 10 \cdot 3 \cdot 4\\ &= 10 \cdot 12\\ &=\textcolor{#4257b2}{120} \end{align*}

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