jakolik wrote:
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)
Bunuel wrote:
jakolik 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 if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
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.
Committee has following formations-
M----------W
2----------4
3----------3
Quote:
Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)
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\(= 700-65=635\)