From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. How many different committees are possible if

(a) 2 of the men refuse to serve together?

(b) 2 of the women refuse to serve together?

(c) 1 man and 1 woman refuse to serve together?

From a

jeffp

Answered by Aaron 11 months ago

**part (a)**

For the men, we choose 3 from 6 order not mattering,

$$\begin{pmatrix} 6 \\ 3 \end{pmatrix} = 20$$

but we need to subtract from this the number of orderings where the two men serve together. This is given by,

$$\begin{pmatrix} 2 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \end{pmatrix} = 4$$

so we have,

$$20 - 4=16$$ men to choose from.

There are $ \begin{pmatrix} 8 \\ 3 \end{pmatrix} = 56 $ different groups of women. So the total number of committees possible is,

$$16 \cdot 56 = 896$$

**part (b)**

This is similar to **part (a)** so we can jump straight to,

$$\bigg[ \begin{pmatrix} 8 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \end{pmatrix} \begin{pmatrix} 6 \\ 1 \end{pmatrix} \bigg] \begin{pmatrix} 6 \\ 3 \end{pmatrix} = 50\cdot 20 = 1000$$ different committees.

**part (c)**
Total number of committees that could possibly exist are,

$$ \begin{pmatrix} 8 \\ 3 \end{pmatrix} \begin{pmatrix} 6 \\ 3 \end{pmatrix} =1120$$

the number of committees that exist where the man and women serve together is given by,

$$ \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 7 \\ 2 \end{pmatrix} = 210$$

so the total number of committees in this case amounts to,

$$1120 - 210 = 910$$

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Asked: 11 months ago