Question

In how many 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.?

Answer 1

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:

Given : 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

To find : In how many ways the given situation satisfied?

Solution :

A team of 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.

There are following number of cases which are

• When No men is selected

$^5C_0\times ^{11}C_{11}=1\times 1=1$

• When  1  men is selected

$^5C_1\times ^{11}C_{10}=5\times 11=55$

• When 2 men is selected

$^5C_2\times ^{11}C_{9}=10\times 55=550$

• When 3 men is selected

$^5C_3\times ^{11}C_{8}=10\times 165=1650$

Total number of ways are

$1+55+550+1650=2256$

Therefore, 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.