Question

The number of ways in which $10$ persons can go in two cars so that there may be $5$ in each car, supposing that two particular persons will not go in the same car is:

• A. $\frac{1}{2}{}^{10}C_5$
• B. $\frac{1}{2}{}^{8}C_5$
• C. $2{}^{8}C_4$
• D. ${}^{8}C_4$

Answer 1

The correct option is C

$2{}^{8}C_4$

Explanation for the correct option:

Find the required number of permutations.

We need to arrange $10$ persons in two cars such that there are $5$ persons in each car and two particular persons will not go in the same car.

So, two persons can be arranged in two cars in $2$ different ways.

Now, we have to arrange $8$ persons in two cars.

So, the number of ways in which four persons can be selected out of $8$ persons is ${}^{8}C_4$.

Therefore, the total number of ways in which $10$ persons can go in two cars so that there may be $5$ in each car, supposing that two particular persons will not go in the same car is $2{}^{8}C_4$.

Hence, option $C$ is the correct answer.