Q. 1.11 - A First Course in Probability

Question

From $10$ married couples, we want to select a group of $6$ people that is not allowed to contain a married couple.

(a) How many choices are there?

(b) How many choices are there if the group must also consist of $3$ men and $3$ women?

Step-by-Step Solution

Answer

(a) The no. of choices are $13440$.

(b) The no. of choices are $4200$.

Part (a) Step 1. Given information.

Total no. of married couples $=10$

No. of people to be selected $=6$ who are not married couple.

Part (a) Step 2. Find the no. of choices.

Total no. of persons $=10\times 2=20$

From these $20$ people, we have to select $6$ such people who are not married couple.

The first person is to be selected out $20$ persons, so the no. of ways will be $20$.

Now as the selected persons cannot be married couple, so the second person will be selected from $18$ persons (excluding the partner of the first person who was selected).

So, the second person can be selected in $18 \text{ways}$.

Similarly, the no. of ways in which six persons can be selected is $20\times 18\times 16\times 14\times 12\times 10$

These six persons can be arranged in $6! \text{ways}$.

Therefore, the total no. of choices are $\frac{20\times 18\times 16\times 14\times 12\times 10}{6!}=13440$.

Part (b) Step 3. Find the no. of choices if the group must also consist of 3 men and 3 women.

From $10$ couples, $3$ men can be selected in $\left(\begin{array}{c}10\\ 3\end{array}\right)$ ways.

Now as the selected persons cannot be married couple, so the $3$ women will be selected from the remaining $7$ couples in $\left(\begin{array}{c}7\\ 3\end{array}\right)$ ways.

Therefore, the total no. of choices are $\frac{10!}{3!7!}\times \frac{7!}{3!4!}=4200$.