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

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 in 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

Total No. of possible combinations possible = (3 Men, 2 Women) + (4 Men, 1 Women) + (5 Men, 0 Women)
= (7C3 * 6C2) + (7C4*6C1) + (7C5)
= 756

Hence, correct answer should be D.

Firstly the formula would be 7C3*10C2, but that too would be wrong here because you will be calculating repetition too.

The men chosen in certain combinations will turn out to be the same ratio n 7C3 and 10C2.

Say abc and de
Next it could be abd and ce.
So repetition

Solution:

At least 3 men in the question implies 3men and 2 women Or 4 men and 1 woman Or 5 men and 0 women.

The Or condition implies an addition here after every case is computed.

Hence

(7C3)*(6C2) to select 3men and 2 women or

(7C4)*(6C1) to select 4 men and 1 woman or

(7C5)*(6C0) to select 5 men and 0 women.

Adding all we have (7C3)*(6C2) + (7C4)*(6C1) + (7C5)*(6C0)

= 525 +210+ 21

= 756 (Option D)

Hope this helps
Devmitra Sen(Quants GMAT Expert)

Probability (at least 3 men) = prob(3 men, 2 women) + prob(4 men, 1 woman) + prob(5 men) --> (7C3)(6C2)+(7C4)(6C1)+(7C5) = (15)(35)+(6)(35)+21 = 756

D.

Can this be done in a way where we select total and then subtract the "anti-case"

7c3*6c2=525
7c4*6c1=210
7c5=21
sum = 756
IMO D

Hi!

Could you please explain what my solution assumes and what exactly it means?

At least 3 men: 7C3 = 35 combinations
2 people from the left ones: 10C2 (10 = 4 men + 6 women are left) = 45 combinations

35 * 45 = 1575

Thank you.

Anyone please ...

Can we not do 7C3 for at least 3 men in the committee and for remaining 2 committee members that we can select from 2 men & 6 women i.e. 8C2

Therefore 7C3 X 8C2 ...What's wrong in this ? Please

SagarV, finding it hard to say what is wrong with your approach in the mathematical sense. But logically, you're considering men=women in your last scenario. I don't think that is correct and is hence is bound to give you the incorrect answer.

The challenge with this question is that there is no anti-case shortcut.

For both the 'main case' and the 'anti-case', you are calculating the sum of 3 parts. Let me try to spell it out below:

Main case:

Sum of combinations in Scenario 1 (3 men, 2 women), Scenario 2 (4 men, 1 woman), and Scenario 3 (5 men, zero women)

Anti-case

Sum of combinations in Scenario 1 (zero men, 5 women), Scenario 2 (1 man, 4 women), and Scenario 3 (2 men, 3 women)

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