A committee of 6 is chosen from 8 men and 5 women so as to

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\)

700-65 = 635

Answer: E.
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Denote by A and B the two particular men.
We can have 2 men and 4 women chosen, when the two men are the particular two: choose the two men \(C^2_2\), then choose 4 women out of the 5, which is \(C^4_5\). So, A,B and four women chosen.
We can also have the two men chosen, again \(C^2_2\), choose another man from the remaining 6 men, this is \(C^1_6\), then choose 3 women out of the 5, which is \(C^3_5.\) Here, A, B, another men, and 3 women are chosen.
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Let M = men and W = women
A committee of 6 is to be chosen.
At least 2M and 3W should be there.
So we can have either 2M+4W OR 3M+3W.

Total possibilities = (8C2)*(5C4) + (8C3)*(5C3)
= 700.

But we are told that two men refuse to serve together.

For, 2M+4W we can select 2M from only 2 men because two men refuse to serve together.
So, (2C2)*(5C4) = 5.

For, 3M+3W we can select 2M from only 2 men because two men refuse to serve together and the other 1M can be selected from remaining 6M.
= 2M + 1M + 3W
= (2C2)*(6C1)*(5C3) = 60.
Hence total = 5 + 60 = 65.

Now, required possibilities = 700 - 65 = 635
Hence option (E).

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Committee has following formations-
M----------W
2----------4
3----------3



This value 700 includes committees in which both of the men (say X and Y) are included. So, we need to find out number of such committees and deduct those from the total 700.

In the first case, 2 men are chosen. No. of committees of 2 men in which X and Y are included is 1
So, Considering women also we find total such committees\(= 1 * ^5C_4\) \(= 5\)

In the second case, 3 men are chosen out of 6.
Also, out of 3 men X and Y will always be in the committee. Therefore, we need to find out 1 more man out of remaining 6. This can be done \(^6C_1\)way = 6 ways

So, Considering women also we find total such committees\(= 6 * ^5C_3\) \(= 6*10\)

Therefore, total no. of committees in which both X and Y are included is 5+60=65

This gives us answer, total no. of committees in which both X and Y are not included\(= 700-65=635\)

Find out how many different combinations can be made between the 8 men and 5 women ...

We will have two scenarios , First where we have 2 men and 4 women , and second where we have 3 men and 4 women ... Finding the combinations for each and then adding them up gives us 700 ...

Logically the number of combinations where any 2 men do not serve together would have to be lower than 700 , therefore the only option is E ....
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Bunuel's subtraction way works. Here's another method -

There are 2 men and 3 women leaving out one spot.

Case 1: We have 3 men and 3 women. Total number of ways = (8C3 - 6)* 5C3 = 500
Case 2: We have 2 men and 4 women. Total number of ways = (8C2 - 1)*5C4 = 135

Total = 635

I didn't see this until I answered the question, but this is a quicker way to get to the answer.

First case 2 men and 4 Women

Let M1 and M2 refuse to serve together

When M1 serves and not M2 , number of ways 6C1*5C4=30
Similarly 30 ways when M2 and not M1 serves
When both M1 nd M2 do not serve 6C2*5C4=75
Total number of ways = 135

Second case of 3 men and 3 women

Using the above reasoning number of ways is 6C2*5C3 + 6C2*5C3 + 6C3*5C3 = 500

Total = 135+500=635
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Thank you Bunuel.

Can you please explain why ways to chose 6 members committee without restriction cannot be: C^2_8*C^3_5*C^1_8
We are to choose 2 men out of 8, 3 women out of 5 and 1 (woman or man) out of 8 (13-5)

This way there will be duplicates in the combinations.

Consider M1,M2....M8 be men and W1,W2......W5 be women
See the below 3 combinations. They are in fact same but will be counted as different with the approach you mentioned

...2M.. .....3W...... 1M or 1W
M1,M2| W1,W2,W3| M3
M3,M2| W1,W2,W3| M1
M3,M1| W1,W2,W3| M2

Thanks

Bunuel, can you explain the logic behind this?

Yups Like others I too have the same query

Anybody explain the logic selecting two men who work together with each other , please?

We only have the symbols here but nothing explaining the same ?
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Hi,
I went at it a different way and did not get the right result.
Can you find out where the logic is wrong?

Total number of possibilities: choose 2 men, choose 3 women, choose one from the leftover: 8c2*5c3*8c1
Total number of possibilities for the 2 men together: 2C2*5C3*8C1.
And last step, subtract the two numbers.... This is not working for me... Why?

I understand that Bunuels way works.... My question is why my way doesn't?

Check here: 4-professors-and-6-students-are-being-considered-for-85865-20.html#p1265096
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spent some time solving this bad boy..
so we can have 2 options:
2M and 4W
and
3M and 3W.

8C2*5C4 + 8C3*5C3 = 700.
this is without restrictions. logically, we can eliminate all answer choices except for E, which is less than 700. and it has to be 700,if we apply restrictions.

nevertheless, with restrictions, I found it difficult to solve...
tried to combine the 2 guys as 1 person:
7C1 * 5C4 = 35
7C2 * 5C3 = 210
245 total ways..but this is way too much...

Hi Brunel, I understood process of finding solution with total-unaccepted solution

But before looking your explanation I started solving question in below manner-


1)2M and 4W
we have restriction that both person should not be together,one of them can be appear in commitee though.

so lets say A & B shouldnt be together
a)when one of them is present lets say A is present and B is not we have 7C2*5C4
b)when both A and B not selected we have option as 6C2*5C4
a+b=total ways of forming committee=180

2)3M3W

a)when one of them is present lets say A is present and B is not we have 7C3*5C3
b)when both A and B not selected we have option as 6C3*5C3
a+b=total ways of forming committee=550

so TOTAL=180+550=730

let me know please which combination I included twice..why I am getting wrong ans

Thanks in advance

Hi,
the method you have adopted is longer and prone to error..
It is much simpler to find ways both are together and subtract from total to get the answer..
lets see the error in your solution..


the error is in coloured portion..
1) 7C2 means selecting 2 out of 7, so this is including ways where even A is not included..
2) It just means all the way where B is not there..
3) So your equation where A is surely there and B is not there will be 6C1*5C4, as one of the men is already occupied by A = 30,..
4) the answer of 3) should be doubled, as it would give us where B is there and A is not there =30*2=60..
5) total ways would be 6C2*5C4+60=135..



I'll leave it for you to do in the similar way as the ways for 2M4W..
it should get you 500



hope you are clear where you went wrong..
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Thanks a ton! Now i understood my mistake