An examination paper consists of 5 questions in section A and 5 questions in section B. A total of 8 questions must be answered. In how many ways can a student select the questions if he is to answer at least 4 questions from section A.

Student has to solve at least $4$ questions from section A.

So, he can choose in 2 ways :

$a)$ $4$ from section A and $4$ from section B.

$b)$ $5$ from section A and $3$ from section B.

Total number of ways of selecting the questions $N=$ $N_a+N_b$

$\ N=^5C_4\times ^5C_4 + ^5C_5\times ^5C_3$

$N=\frac{5!}{(5-4)!(4!)}\times \frac{5!}{(5-4)!(4!)}+\frac{5!}{(5-5)!(5!)}\times \frac{5!}{(5-3)!(3!)}$

$N=\frac{5!}{1!4!}\frac{5!}{1!4!}+\frac{5!}{0!5!}\frac{5!}{2!3!}=5\times 5+ 1\times 10$

$N=25+10=35$