Here's the question you clicked on:
abs4
A committee of 5 people is chosen from 6 men and 4 women. in how many ways can this be done 1) if there must be more men than women on the commitee. 2) if there must be 3 men and 2 women, and one particular woman refuses to be on the commitee with one particular man
Permutations and combinations....
More men is when No. of men is 3 or No. of men is 4 or No. of men is 5
Which means there must be atleast three men
6C3*7C2 First we choose at least three 3 men then choose any two people from the remaining people
the marking scheme says choose 4 men and 1 women or 5 men and no women. i don't understand why we can't choose 3 men and 2 women.
I mean u can choose 3 men and 2 women
yeah but i wonder why in my mark scheme it's not given. that's why the total answer is 186 but i got 180. by choosing 4 men, 1 women and 3 men 2 women. thanks anyways..:)
do you know how to do the second part?
6C3 * 4C2 + 6C4 *4C1 +6C5 =120+60+6=186
U forgot the case when all are men
Can someone tell me why 6C3 *7C2 doest work
I was also wondering the same thing. hmm..
M1 M2 M3 M4 M5 M6 W1 W2 W3 W4 Can we try to rationalize ?
@sauravshakya, can you explain how u got 6C3*7C2 ?
First we choose at least three 3 men then choose any two people from the remaining people