Question

A student has to answer $10$ questions, choosing at least $4$ questions from each of parts $A$ and $B$. If there are $6$ questions in part $A$ and $7$ questions in part $B$, in how many ways can the student choose $10$ questions?

Answer 1

Total no. of questions in part A $=6$

Total no. of questions in part B $=7$

A student can choose $10$ questions from the given $13$ questions in the following pattern-

1. $4$ from part A and $6$ from part B
2. $5$ from part A and $5$ from part B
3. $6$ from part A and $4$ from part B

$\therefore$ Total no. of ways can the student choose $10$ questions $=({}_6{C}_4\times {}_7{C}_6)+({}_6{C}_5\times {}_5{C}_5)+({}_6{C}_6\times {}_7{C}_4)$
As we know that,

${}_n{C}_r=\frac{n!}{r!(n-r)!}$

$\therefore$ total no. of ways can the student choose $10$ questions $=(15\times 7)+(6\times 21)+(1\times 35)=266$

Hence the no. of ways can the student choose $10$ questions are $266$.

Hence the correct answer is $266$.