Bunuel wrote:
Geronimo 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?
(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510
(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510
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.
Been trying to figure this one out, but I'm not getting your calculations when choosing it with the two men together. How do you arrive at C^2_2 and the (C^2_2*C^1_6) . Sorry if I'm being a bit slow on this one, but just haven't wrapped my brain around this.