Question

In a Mathematics paper there are three sections containing 4, 5 and 6 questions respectively. From each section 3 questions are to be answered. In how many ways can the selection of questions be made ?
(A) 34 (B) 800 (C) 1600 (D) 9600​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, In a Mathematics paper, there are three sections containing 4, 5 and 6 questions respectively. From each section 3 questions are to be answered.

So, Number of ways in which 3 question out of 4 can be answered from first section is

$\rm \: = \: ^{4}C_{3}$

$\rm \: = \: \dfrac{4!}{3! \: (4 - 3)!}$

$\rm \: = \: \dfrac{4 \times 3!}{3! \: 1!}$

$\rm \: = \: 4$

Now, Number of ways in which 3 question out of 5 can be answered from second section is

$\rm \: = \: ^{5}C_{3}$

$\rm \: = \: \dfrac{5!}{3! \: (5 - 3)!}$

$\rm \: = \: \dfrac{5 \times 4 \times 3!}{3! \: 2!}$

$\rm \: = \: 10$

Now, Number of ways in which 3 question out of 6 can be answered from third section is

$\rm \: = \: ^{6}C_{3}$

$\rm \: = \: \dfrac{6!}{3! \: (6 - 3)!}$

$\rm \: = \: \dfrac{6 \times 5 \times 4 \times 3!}{3! \: 3!}$

$\rm \: = \: \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1}$

$\rm \: = \: 20$

So, Number of ways in which selection can be made is

$\rm \: = \: 4 \times 10 \times 20$

$\rm \: = \: 800$

Hence, Option (B) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:^{n}C_{r} \: = \: ^{n}C_{n - r} \: }}$

$\boxed{ \rm{ \:^{n}C_{r} \: = \: \frac{n}{r} \: ^{n - 1}C_{ r - 1} \: }}$

$\boxed{ \rm{ \:^{n}C_{r} \: + \: ^{n}C_{r - 1} \: = \: ^{n + 1}C_{r} \: \: }}$

$\boxed{ \rm{ \: \frac{^{n}C_{r}}{^{n}C_{r - 1}} = \frac{n - r + 1}{r} \: \: }}$