## abs4 2 years ago 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

1. nitz

Permutations and combinations....

2. sauravshakya

More men is when No. of men is 3 or No. of men is 4 or No. of men is 5

3. sauravshakya

Which means there must be atleast three men

4. sauravshakya

6C3*7C2 First we choose at least three 3 men then choose any two people from the remaining people

5. sauravshakya

got it?

6. abs4

the answer is 186

7. abs4

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.

8. sauravshakya

u can do it too

9. sauravshakya

I mean u can choose 3 men and 2 women

10. abs4

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..:)

11. abs4

do you know how to do the second part?

12. sauravshakya

6C3 * 4C2 + 6C4 *4C1 +6C5 =120+60+6=186

13. sauravshakya

U forgot the case when all are men

14. sauravshakya

Can someone tell me why 6C3 *7C2 doest work

15. shubhamsrg

I was also wondering the same thing. hmm..

16. shubhamsrg

M1 M2 M3 M4 M5 M6 W1 W2 W3 W4 Can we try to rationalize ?

17. saloniiigupta95

@sauravshakya, can you explain how u got 6C3*7C2 ?

18. sauravshakya

First we choose at least three 3 men then choose any two people from the remaining people