Question

Previous year Question of jee mains

mathamatics shift :1
24 feb 2021​

Answer 1

Answer:

Answer in the attachment.

Answer 2

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

Given that

➢ There are 6 Indians and 8 foreigners.

➢ Now, we have to find the number of committees with atleast 2 Indians and number of foreigners is twice the number of Indians.

➢ So, Following three cases arises :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Indians & \bf foreigners \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 & \sf 4 \\ \\ \sf 3 & \sf 6 \\ \\ \sf 4 & \sf 8 \end{array}} \\ \end{gathered}$

So,

1. Number of ways in which 2 Indians and 4 foreigners can be selected from 6 Indians and 8 foreigners is

$\rm \:  =  \:  \:^{6}C_{2} \: \times \: ^{8}C_{4}$

$\rm \:  =  \:  \:\dfrac{6!}{2! \: (6 - 2)!} \times \dfrac{8!}{4! \: (8 - 4)!}$

$\rm \:  =  \:  \:\dfrac{6 \times 5 \times 4!}{2 \times 1 \times \: 4!} \times \dfrac{8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \: \times 4!}$

$\rm \:  =  \:  \:15 \times 70$

$\rm \:  =  \:  \:1050$

2. Number of ways in which 3 Indians and 6 foreigners can be selected from 6 Indians and 8 foreigners is

$\rm \:  =  \:  \:^{6}C_{3} \: \times \: ^{8}C_{6}$

$\rm \:  =  \:  \:\dfrac{6!}{3! \: (6 - 3)!} \times \dfrac{8!}{6! \: (8 - 6)!}$

$\rm \:  =  \:  \:\dfrac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times \:3!} \times \dfrac{8 \times 7 \times 6!}{6! \times \: 2 \times 1}$

$\rm \:  =  \:  \:20 \times 28$

$\rm \:  =  \:  \:560$

3. Number of ways in which 4 Indians and 8 foreigners can be selected from 6 Indians and 8 foreigners is

$\rm \:  =  \:  \:^{6}C_{4} \: \times \: ^{8}C_{8}$

$\rm \:  =  \:  \:\dfrac{6!}{4! \: (6 - 4)!} \times \dfrac{8!}{8! \: (8 - 8)!}$

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

$\rm \:  =  \:  \:15$

Hence,

➢ Total number of ways in which the number of committees with atleast 2 Indians and number of foreigners is twice the number of Indians is

$\rm \:  =  \:  \:1050 + 560 + 15$

$\bf \:  =  \:  \:1625$

Hence, Option (1) is correct

Additional Information :-

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

$\boxed{ \sf{ \:^nC_ r \: + \: ^nC_ {r - 1} = \: ^{n + 1}C_r}}$

$\boxed{ \sf{ \:\dfrac{^nC_ r}{^nC_{r - 1}} = \dfrac{n - r + 1}{r}}}$

$\boxed{ \sf{ \:^nC_ 0 \: = \: ^nC_ n \: = 1}}$

$\boxed{ \sf{ \:^nC_ r = \frac{n!}{r! \: (n - r)!}}}$