How can I solve this question by using Hyper geometric law ?

[Bel]

In a club of six men and six women, in how many ways can a committee of four be selected if : 1-The committee must consist of two men and two women 2-There must be at least one of each sex
3-The women must outnumber the men ?

 

[pka]

In a club of six men and six women, in how many ways can a committee of four be selected if :
1-The committee must consist of two men and two women
2-There must be at least one of each sex
3-The women must outnumber the men ?

Before posting, did you read the RULES FOR POSTING? If you had the you know to post some of your own work.
So do you understand that there are \(\displaystyle \dbinom{12}{4}\) total possible committees? If so explain it to us.
If not then tell us why not.

 

[Selim47]

Are there 3 different questions? If not, isn't there a conflict between condition no 1 and 3 ?

 

[pka]

Are there 3 different questions? If not, isn't there a conflict between condition no 1 and 3 ?

It is usual to number separate questions as above. On there other hand, using sub-questions a), b) etc. can imply interconnections( but not always).

 

[Dr.Peterson]

In a club of six men and six women, in how many ways can a committee of four be selected if :
1-The committee must consist of two men and two women
2-There must be at least one of each sex
3-The women must outnumber the men ?

Your title mentions the "hypergeometric law". What have you learned about that? Please state the formula as you learned it, and the meaning of each variable, so we can help you use it.

Have you tried applying it to the three problems here? Or are you not required to use that specific formula, and can just think through the problems with combinations?

 

[Steven G]

Are there 3 different questions? If not, isn't there a conflict between condition no 1 and 3 ?

Yes, they are 3 different questions.

1 and 2 are different. 1 says EXACTLY 2 men and 2 women. Now 2 says at least 1 of each sex which can be 1 man, 3 women or 2 men, 2 women or 3 men, 1 women. This is very different from 1.

2 and 3 are different. I just said what 2 means. 3 means that there can be 3 women, 1 man OR 4 women, 0 men. Again, this is different from 2

1 and 3 are different. See above