91Ó°ÊÓ

It is given that,

Total no. of questions $=10$

No. of questions to be answered $=7$

A student can select $7$out of $10$questions in $\left(\begin{array}{c}10\\ 7\end{array}\right)\text{ways}$

$=\frac{10!}{7!3!}=\frac{10×9×8×7!}{3×2×1×7!}=5×3×8=120$

Therefore, the no. of choices for selecting$7$out of$10$questions is$120$.

Let us divide the total questions in two groups, each group containing $5$questions.

If a student selects $3$questions from first group then he will select $4$questions from the second group. So, the no. of choices will be

$$\begin{gathered} \left(\begin{array}{c}5\\ 3\end{array}\right)\left(\begin{array}{c}5\\ 4\end{array}\right)=\frac{5!}{3!2!}×\frac{5!}{4!1!} \\ =10×5 \\ =50 \end{gathered}$$

If a student selects $4$questions from first group then he will select $3$questions from the second group. So, the no. of choices will be

role="math" localid="1648646922678" $$\begin{gathered} \left(\begin{array}{c}5\\ 4\end{array}\right)\left(\begin{array}{c}5\\ 3\end{array}\right)=\frac{5!}{4!1!}×\frac{5!}{3!2!} \\ =5×10 \\ =50 \end{gathered}$$

If a student selects $5$questions from first group then he will select $2$questions from the second group. So, the no. of choices will be

$$\begin{gathered} \left(\begin{array}{c}5\\ 5\end{array}\right)\left(\begin{array}{c}5\\ 2\end{array}\right)=\frac{5!}{5!}×\frac{5!}{3!2!} \\ =1×10 \\ =10 \end{gathered}$$

Therefore, the no. of choices of selecting $7$out of $10$questions, if at least $3$questions are to be selected from first $5$ questions$=50+50+10=110$