Possibilities:800guy 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?
is the answer 560+140= 700?800guy 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?
I am getting 665. Why are you subtracting 6 in the second case? It should be 3 according to me.Please explain.ThanksOA:
4W 2M == 5C4.(8C2-1) = 5.(27) = 135
3W 3M == 5C3.(8C3-6) = 10.50 = 500
total 635
Nice question.. I have a query here.. in the bold part above if 2 men are always together then why don't u subtract those 2 from 6 and then select 1 out of remaining 4.. so we have those 2 already + 1 from the remaining 4 men.. so we can select one of them in 4C1 ways.800guy wrote:OA:
4W 2M == 5C4.(8C2-1) = 5.(27) = 135
3W 3M == 5C3.(8C3-6) = 10.50 = 500
total 635
Max. number of possibilities considering we can choose any man 8c2 * 5C4 + 8C3*5c3 = 700.
consider it this way.... from my previous reply max possible ways considering we can chose any man = 700
now we know that 2 man could not be together... now think opposite... how many ways are possible to have these two man always chosen together...
since they are always chosen together...
For chosing 2 men and 4 women for the committee there is only 1 way of chosing 2 men for the committee since we know only two specific have to be chosen and there are 5C4 ways of choosing women
1*5C4 = 5
For choosing 3 men and 3 women for the committee there are exactly 6C1 ways choosing 3 men for the committee since we know two specific have to be chosen so from the remaining 6 men we have to chose 1 and there are 5C4 ways of choosing women
6C1*5C3 = 60
So total number of unfavorable cases = 5 + 60 = 65
Now since we want to exclude these 65 cases... final answer is 700-65 = 635
