Problem Solving

Hey, I'm looking to see if anyone can give me some insight into this problem; it won't be verbatim but close enough:

Out of a group of 12 temporary employees, 4 will be hired. 5 out of the 12 temporary employees are women. How many of the possible four person groups will consist of 3 women and 1 man?

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I keep getting 35/2, while the practice test says the correct answer is 70. I can't get where they are getting this factor of four!!

I found the total number of groups via: 12!/(4!*8!)

I found the odds of three woman and one man as: (5/12)(4/11)(3/10)(7/9)

I multiply the two and then get 35/2.

Any help would be appreciated!!

Cheers,
Aashish

PS: My screen name existed way before the recent blockbuster hit!!

well it goes like this....
3 womens from 5 have to b selectes so it becomes 5C3= 10
similarly 1 man from reamining 12-5=7 men avail. so, 7C1=7

so total number of ways to choose is 10*7=70

it hope its clear!!

ok, i got 70 as well but my question is, why would we not multiply the 70 times 12C4 since that would tell her the group combination possibilities?

well 12C4 gives the possibility of picking any 4 ppl from group of 12...but the que. specifically segregates the group in man and women.
so man could be choosen only from avail. number, similarly women also.

total combination will be mulitpication of both.

i hope its clear!!!

ok, so we're including the 12C4 since 7C1 and 5C3 assume that there are 12 people and four are being chosen.

got it!

thanks rammy :0)

So you compute the woman's total combinations and then the men's and then multiply them?

I can see how that works, and how that makes things easier, but why doesn't the method I tried work? I don't understand why am not able to use my first general method.. I guess I'm missing something in the concept.

Thanks already for the help, any further help would be appreciated!

Aashish

Any instructors out there know why my first strategy doesn't work??

Thanks again,
Aashish

Thanks, Ian. That definitely makes sense. I just assumed since the odds of picking the man either last or first, etc. were the same, it would have been sufficient. But your explanation was great.

Thanks again,
Aashish