From a group of 7 men and 6 women, five persons are to be...

[BTGmoderatorLU]

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A) 564
B) 645
C) 735
D) 756
E) 566

The OA is D.

I'm confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

[GMATWisdom]

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A) 564
B) 645
C) 735
D) 756
E) 566

The OA is D.

I'm confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

WE CAN SELECT A COMMITTEE OF 5 PERSONS WITH AT LEAST 3 MEN IN EACH GROUP IN THE FOLLOWING WAYS:-
1. 3 MEN AND 2 WOMEN . THIS WILL LEAD TO NUMBER OF GROUPS = 7C3 * 6C2
2. 4 MEN AND 1 WOMAN. THIS WILL LEAD TO NUMBER OF GROUPS==7C4*6C1
3. 5 MEN . THIS WILL LEAD TO NUMBER OF GROUPS =7C5
THEREFORE TOTAL NUMBER OF GROUPS = 7C3 * 6C2 +7C4*6C1+ 7C5 = 35X15 +35X6+21=525+210+21=756
HENCE OPTION " D"

[Scott@TargetTestPrep]

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A) 564
B) 645
C) 735
D) 756
E) 566

The OA is D.

I'm confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

We are given a group of 7 men and 6 women and need to determine the number of ways a committee of 5 can be formed with at least 3 men on the committee.

Thus, we have 3 scenarios in which at least 3 men can be selected for the 5-person committee: 3 men and 2 women OR 4 men and 1 woman OR 5 men. Let's calculate the number of ways to select the committee for each scenario.

Scenario 1: 3 men and 2 women

Number of ways to select 3 men: 7C3 = (7 x 6 x 5)/3! = (7 x 6 x 5)/(3 x 2 x 1) = 35

Number of ways to select 2 women: 6C2 = (6 x 5)/2! = (6 x 5)/(2 x 1) = 15

Thus, the number of ways to select 3 men and 2 women is 35 x 15 = 525.

Scenario 2: 4 men and 1 woman

Number of ways to select 4 men: 7C4 = (7 x 6 x 5 x 4)/4! = (7 x 6 x 5 x 4)/(4 x 3 x 2 x 1) = 35

Number of ways to select 1 woman: 6C1 = 6

Thus, the number of ways to select 4 men and 1 woman is 35 x 6 = 210.

Scenario 3: 5 men

Number of ways to select 5 men: 7C5 = (7 x 6 x 5 x 4 x 3)/5! = (7 x 6 x 5 x 4 x 3)/(5 x 4 x 3 x 2 x 1) = 42/2 = 21

Thus, the number of ways to select a 5-person committee with at least 3 men is:

525 + 210 + 21 = 756

Answer: D