swerve wrote:A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

A. 635
B. 700
C. 1404
D. 2620
E. 3510

Since the committee must have 6 people, there are two cases that meet the given restrictions:
1) The committee has 2 men and 4 women
2) The committee has 3 men and 3 women

1) The committee has 2 men and 4 women
Since the order in which we select the men and women does not matter, we can use COMBINATIONS
We can select 2 men from 8 men in 8C2 ways (= 28 ways)
We can select 4 women from 5 women in 5C4 ways (= 5 ways)
So, the total number of ways to select 2 men and 4 women = 28 x 5 = 140

2) The committee has 3 men and 3 women
We can select 3 men from 8 men in 8C3 ways (= 56 ways)
We can select 3 women from 5 women in 5C3 ways (= 10 ways)
So, the total number of ways to select 2 men and 4 women = 56 x 10 = 560

So, the TOTAL number of ways to create the 6-person committee = 140 + 560 = 700

Answer: B

Cheers,
Brent

swerve wrote:A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

A. 635
B. 700
C. 1404
D. 2620
E. 3510

The OA is B

Source: Economist GMAT

Since the committee of 6 must contain at least 2 men and 3 women, we can have two scenarios:

1) 2 men and 4 women, or 2) 3 men and 3 women

Scenario 1): 2 men and 4 women.

2 men:

8C2 = (8 x 7)/2! = 28

4 women:

5C4 = 5

The committee of 2 men and 4 women can be selected in 28 x 5 = 140 ways.

Scenario 2): 3 men and 3 women.

3 men:

8C3 = (8 x 7 x 6)/3! = 56

3 women:

5C3 = (5 x 4 x 3)/3! = 10

The committee of 3 men and 3 women can be selected in 56 x 10 = 560 ways

So the total number of ways to select a committee is 140 + 560 = 700.

Answer: B

swerve wrote:A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

A. 635
B. 700
C. 1404
D. 2620
E. 3510

The OA is B

Source: Economist GMAT

Since the committee of 6 must contain at least 2 men and 3 women, we can have two scenarios:

1) 2 men and 4 women, or 2) 3 men and 3 women

Scenario 1): 2 men and 4 women.

2 men:

8C2 = (8 x 7)/2! = 28

4 women:

5C4 = 5

The committee of 2 men and 4 women can be selected in 28 x 5 = 140 ways.

Scenario 2): 3 men and 3 women.

3 men:

8C3 = (8 x 7 x 6)/3! = 56

3 women:

5C3 = (5 x 4 x 3)/3! = 10

The committee of 3 men and 3 women can be selected in 56 x 10 = 560 ways

So the total number of ways to select a committee is 140 + 560 = 700.

Answer: B