Question

N 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

8+3=11 formed one group and the other group I don't know

Answer 2

Answer:

2256  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.

Step-by-step explanation:

To find : 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?

Solution :

Number of men in team = 5

Number of women in team = 11

Maximum 3 men is selected and we have to choose 11 which means there can be 0, 1, 2, 3 men in the team.

Number of cases are

1) No men is selected $^5C_0\times ^{11}C_{11}$

2) 1  men is selected $^5C_1\times ^{11}C_{10}$

3) 2 men is selected $^5C_2\times ^{11}C_{9}$

4) 3 men is selected $^5C_3\times ^{11}C_{8}$

Total ways are

$=^5C_0\times ^{11}C_{11}+^5C_1\times ^{11}C_{10}+^5C_2\times ^{11}C_{9}+^5C_3\times ^{11}C_{8}$

$=1\times 1+5\times 11+10\times 55+10\times 165$

$=1+55+550+1650$

$=2256$

Therefore, 2256  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.