Q. 1.20 - A First Course in Probability

Question

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?

Step-by-Step Solution

Answer

(a) The number of choices are $36$. 

(b) The number of choices are $26$.

Part (a) Step 1. Given information.

It is given that,

Total number of friends $=8$

No. of friends to be invited to a party  $=5$

Part (a) Step 2. Find the number of choices if 2 of the friends are feuding and will not attend together.

The choices are $=\left(\begin{array}{c}8\\ 5\end{array}\right)-\left(\begin{array}{c}2\\ 2\end{array}\right)\left(\begin{array}{c}6\\ 3\end{array}\right)$

$$\begin{gathered} =\frac{8!}{5!3!}-\left[\frac{2!}{2!}\times \frac{6!}{3!3!}\right] \\ =\frac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}-\left[1\times \frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}\right] \\ =56-20 \\ =36 \end{gathered}$$

Part (b) Step 3. Find the number of choices if 2 of the friends will only attend together.

The choices are $=\left(\begin{array}{c}2\\ 2\end{array}\right)\left(\begin{array}{c}6\\ 3\end{array}\right)+\left(\begin{array}{c}2\\ 0\end{array}\right)\left(\begin{array}{c}6\\ 5\end{array}\right)$

$$\begin{gathered} =\left[\frac{2!}{2!}\times \frac{6!}{3!3!}\right]+\left[\frac{2!}{2!}\times \frac{6!}{5!1!}\right] \\ =\left[1\times \frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}\right]+\left[1\times \frac{6\times 5!}{5!}\right] \\ =20+6 \\ =26 \end{gathered}$$