Question

In how many ways a team of 11 must be selected from 5 men and 11 women such that the team must comprise of not more than 3 men?

Answer 1

Here i am assuming that you know basics of permutations and combinations. If yes than i have solved the whole question in above image.

Hope it helps.

Answer 2

Answer:

In 2256 ways  a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.

Step-by-step explanation:

We have given that, a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men

We have to find, in how many ways a team is selected?

Solution :

There are 5 men and 11 women.

Maximum 3 men is selected and we have to choose 11.

The ways men be in the team is  0, 1, 2, 3.

We know, $^nC_k=\frac{n!}{k!(n-k)!}$

Follow the cases :

1) When No men is selected

$^5C_0\times ^{11}C_{11}=\frac{5!}{0!\times 5!}\times \frac{11!}{11!(11-11)!}\\\\^5C_0\times ^{11}C_{11} =1\times 1=1$

2) When 1  men is selected

$^5C_1\times ^{11}C_{10}=\frac{5!}{1!\times (5-1)!}\times \frac{11!}{10!(11-10)!}\\\\^5C_1\times ^{11}C_{10}=5\times 11=55$

3)  When 2 men is selected

$^5C_2\times ^{11}C_{9}=\frac{5!}{2!\times (5-2)!}\times \frac{11!}{9!(11-9)!}\\\\^5C_2\times ^{11}C_{9}=10\times 55=550$

4)  When 3 men is selected

$^5C_3\times ^{11}C_{8}=\frac{5!}{3!\times (5-3)!}\times \frac{11!}{8!(11-8)!}\\\\^5C_3\times ^{11}C_{8}=10\times 165=1650$

Total number of ways they choose is given by,

$1+55+550+1650=2256$

Hence, In 2256 ways  a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.