A person has 8 friends, of whom 5 will be invited to a party.

(a) How many choices are there if 2 of the friends
are feuding and will not attend together?

(b) How many choices if 2 of the friends will only
attend together?

A person

jeffp

Answered by Aaron 2 years ago

**part (a)**
The total number of choices will be choosing 5 from the 8 order not mattering minus those groups who include the feuding friends,

$$\begin{pmatrix} 8 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \end{pmatrix} \begin{pmatrix} 6 \\ 3 \end{pmatrix} = 36$$

**part (b)**
If the two friends are not included in the group than we can get $\begin{pmatrix} 6 \\ 5 \end{pmatrix}$ friends. If the two friends are included then we can get $\begin{pmatrix} 2 \\ 2 \end{pmatrix} \begin{pmatrix} 6 \\ 3 \end{pmatrix}$ friends. These are mutually exclusive events for a total of,

$$ \begin{pmatrix} 6 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \end{pmatrix} \begin{pmatrix} 6 \\ 3 \end{pmatrix} = 26$$

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Asked: 2 years ago