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: A.
_________________

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


What are GMAT Club Tests?
25 extra-hard Quant Tests

Find out what's new at GMAT Club - latest features and updates

Karishma I am not able to get the highlighted portion. Can you pls simplyfy for me.

This is what I still need to improve...
Thx for both answers !!

Well, i do not get the 635 but narrowed it down to answer E

Total (favourable) commitees = All possible commitees - restriction

Option 1: M M W W W W (2 Men & 4 Women)
Option 2: M M M W W W (3 Men & 3 Women)
Restriction: 2 out of 8 Men refuse to serve together

Outcomes Option 1: \frac{8!}{2!*6!} * \frac{5!}{4!*1!} = 28 * 5 = 140

Outcomes Option 2: \frac{8!}{3!*5!} * \frac{5!}{3!*2!} = 56 * 10 = 560

Total commitees = 700

Outcomes restriction option 1: 1 (if 2 men refuse to serve together and 2 places are available) * 5 (Outcomes women) = 5
Outcomes restriction option 2: 3 (if 2 men refuse to serve together and 3 places are available) * 10 (Outcomes women = 30

So it should be definitely less than 700 but i guess i made a mistake somewhere. POE E

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 if two of the men refuse to serve together?

Number of ways to choose 2 men = 8C2 = 28
Number of ways to choose 3 women = 5C3 = 10
Total ways to choose 2 men and 3 women = 28 * 10 = 280

Now the remaining committee member can be chosen from the 8 remaining men and women.
Total ways to choose 6 committee members = 280 * 8 = 2240

When the two men who refuse to serve together are selected -
Number of ways to choose 3 women = 5C3 = 10
Remaining 1 committee member can be chosen from the 8 remaining men and women.
Total combinations = 10 * 8 = 80

Total different committees if two of the men refuse to serve together = 2240 - 80 = 2160


I am doing something wrong here, where am i going wrong?


-----------------------

Realize my mistake, i am duplicating count when i multiply by 8.

Thanks Bunuel for explaining..it's clear to me now..

Be careful when you do not have to arrange the people but need to choose multiple times from the same group.

Say, I want to make a team of 2 people from a group of 4 (A, B, C, D).
In how many ways can I do it?
Method1: 4C2 = 6 (AB, BC, CD, AC, AD, BD)
Method 2: If instead, I select 1 and then another, this is how I will do it: 4C1 * 3C1 = 12 (AB, BA, AC, CA...)
In the second case, I have arranged the 2 people in 2 ways. I first select A and then B in one case. I first select B and then A in another case.
But a team doesn't need arrangement, it only needs selection.
Hence, the second method is incorrect.

Similarly, in this question, you need to make a committee i.e. only select, not arrange. When you pick 2 men, 3 women and 1 of the remaining, you are double counting.

When you select 2 men in 8C2 ways, say you select A and B
When you select 3 women in 5C3 ways, say you select X, Y and Z.
When you select the last person, say you select a man C.

Now consider another case:
When you select 2 men in 8C2 ways, say you select A and C
When you select 3 women in 5C3 ways, say you select X, Y and Z.
When you select the last person, say you select a man B.

The committee is same in both the cases (A, B, C, X, Y, Z) but we count them as 2 different cases when we use your method.
Hence, there is double counting in your method. Therefore, you need to do what Bunuel suggested.

For more on P&C, check my blog: http://www.veritasprep.com/blog/categor ... er-wisdom/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

You are absolutely right. Why do you think you are wrong?
Check the expressions, they will give you the correct answer.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.