Question

How many ways can you form committee with 4 men and 11 women, if there are to be at least 4 men and twice as many women as men?

Answer 1

There are only 4 men available and you need atleast 4 men so you need to take all men

Answer 2

Since women must be twice as many as men, you need to choose exactly 8 women

Answer 3

THe problem simplifies to finding number of ways of choosing 4 men and 8 women

Answer 4

\[\large \rm \binom{4}{4} \times \binom{11}{8}\]

Answer 5

Do you also have to add on the number of ways for having a committee with 8,9,10 and 11 women?

Answer 6

Depends on how you interpret the problem

Answer 7

Like you could have a committee of 4 men, 8 or 9 or 10 or 11 women, And another set of committees with 5 men and 10 or 11 women?

Answer 8

Does " twice as many women as men" imply "atleast twice" ?

Answer 9

Yes at least twice as many women as men.

Answer 10

Ohkay, then just add up other combinations too

Answer 11

5 men is impossible because there are only 4 men available

Answer 12

\[\rm \binom{4}{4} \times \left[\binom{11}{8}+ \binom{11}{9}+ \binom{11}{10}+ \binom{11}{11}\right]\]

something like this ?

Answer 13

Yeah but I think since the order doesn't matter wouldn't I just want to add the men to women?

Answer 14

Here multiplication is not about order, you need to have 4 men for sure in the committee. So you don't have any choice... Maybe assume that they are already there in the committee and select women

Answer 15

If you want to think of it in terms of order, try this : first find # of ways to choose four men. after that, for each of this choice, find the number of ways of choosing 8+ women

Answer 16

Ok. That makes sense. I understand how to do it for four men, with the different amounts of women, Would I then add on the # of ways to have 5 men with 10 and 11 women since that is the only other possible formation since there are 6 men to choose from?

Answer 17

have i misread the main question.... it says there are 4 men only hm

Answer 18

I made a mistake writing it. There are suppose to be 6 men, and 11 women. So with the constraint you should be able to have the choice of 5 men as well.

Answer 19

Sorry...

Answer 20

Got you :) yes you can do the same for (5men 10/11women)

Answer 21

6men is not possible right ?

Answer 22

Correct. Thank you so much Ganeshie!

Answer 23

\[\rm \binom{6}{4} \times \left[\binom{11}{8}+ \binom{11}{9}+ \binom{11}{10}+ \binom{11}{11}\right] \\~\\+ \\~\\\rm \binom{6}{5} \times \left[ \binom{11}{10}+ \binom{11}{11}\right]\]