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

Question

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

testprepabc · Accepted Answer

Since at least 3 men must be chosen, we consider all committees which include 3, 4, and 5 men, with 2, 1, and 0 women, respectively. That is, we want to add the number of ways to: a. Choose 3 from 7 men and 2 from 6 women = (7C3)*(6C2) = 35*15 = 525 b. Choose 4 from 7 men and 1 from 6 women = (7C4)*(6C1) = 35*6 = 210 c. Choose 5 from 7 men and 0 from 6 women = (7C5)*(6C0) = 21*1 = 21 a+b+c= 756. Hence option D Hope it helps

chetan2u · Answer

Hi
A quick observation..
Ans will have selection of atleast 3 from 7 men, so the answer has to be a multiple of 7..
So ONLY C and D are left..
A high probability to get your answer correct even if you did not know what to do..

Now TWO ways..
1) Now select 3 from 7 and select next 2 from remaining 6 females Or both from males or one each..
And the solution as mentioned above will be
7C5+7C4*6C1+7C3*6C2=21+35*6+35*15=756..

2) Another way total ways - ways where 3 males are not there..
All females 6C5=6..
All but one 6C4*7C1=15*7=105
All but 2..6C3*7C2= 20*21=420
So ways where 3 males are not there 6+105+420=531..

Total 13C5= 13*12*11*10*9/(2*3*4*5)= 13*11*9=1287..
So remaining ways = 1287-531=756..

D

nitswat · Answer

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.

law258 · Answer

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.

JeffTargetTestPrep · Answer

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

praneet87 · Answer

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

Archit3110 · Answer

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

Render · Answer

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.

SagarV · Answer

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

MBAB123 · Answer

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.

chetan2u · Answer

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

CrackverbalGMAT · Answer

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)

davidmfox · 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)

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards