Question

In a group of $6$ boys and $4$ girls, four children are to be selected. In how many ways can they be selected such that at least one boy should be there

• A. $109$
• B. $128$
• C. $138$
• D. $209$

Answer 1

The correct option is D

$209$

Explanation for the correct answer:

There are $6$ boys and $4$ girls.

Hence, the total number of children is $10$.

The ways of selecting $4$ children from a group of $10$ is given as ${}^{10}{C}_4$

${}^{10}{C}_4=\frac{10!}{4!6!}=210$

Hence, there are $210$ ways in total of selecting $4$children out of $10$

Consider the groups in which there are no boys, implying all $4$ children are girls.

The ways of forming a group of $4$ children with $4$girls is given as ${}^4{C}_4$

${}^4{C}_4=\frac{4!}{4!0!}=1$

Hence, there is only $1$way a group of $4$ children can be formed such that there are no boys in it.

Rest all groups contain at least $1$boy in them

Number of ways at least $1$boy is selected $=$Total number of ways$-$ Number of ways in which no boys are selected

$\Rightarrow$Number of ways at least $1$boy is selected $=$$210-1=209$

Hence there are $209$ ways in which the group will contain at least $1$boy.

Hence, option $\left(D\right)$is the correct answer.