Permutations and combinations

[diracdelta]

Homework Statement

Out of 7 women and 4 men you need to choose delegation.
On how many ways you can choose delegation that consist of:
a) five people - 3 women and 2 men
b)any number of people, but with equal number of men and women
c) five people with at least 3 women
d) five peope where one member is decided to be women

This is my attempt. I don't know d) part, help please :)

$$a) \binom{7}{3}\binom{4}{2}=210\\ b)\binom{7}{1}\binom{4}{1} + \binom{7}{2}\binom{4}{2}+ \binom{7}{3}\binom{4}{3} + \binom{7}{4}\binom{4}{0}=329\\ c) \binom{7}{2}\binom{4}{3} + \binom{7}{3}\binom{4}{2} + \binom{7}{4}\binom{4}{1} + \binom{7}{5}\binom{4}{0}$$

 

[PWiz]

Are you sure the first term in c) is correct? (At least 3 women...)
And does d) say "one woman only", or what you've posted is the entire question part?

 

[diracdelta]

You are right, its accident. first term shouldn't be there. ( expr. c)
d) says that we have determined that one person is definitely women, other 4 can be both men and women.
- So, first we can choose women in 7 ways. but how do i get other four ?

By the way, a) and b) are correct?

 

[PWiz]

The number of women chosen for delegation can vary from 1 to 5 (since 5 people in total are to be selected and we need at least 1 woman to be chosen). The number of men chosen will then be $5-n$ where $n$ is the number of women that are chosen. So you will have something along the lines of "1 woman and 4 men" OR "2 women and 3 men" OR etc. Can you take it from here?

 

[diracdelta]

I see that. I also have to do the same for men?

One more thing. Do I then sum 1 women and 4 men + ... 4 men & 1 women?

 

[PWiz]

I see that. I also have to do the same for men?

I don't get what you mean. "4 men and 1 woman" is the same thing as "1 woman and 4 men". You just have to cycle through different combinations by adding 1 to number of woman and reducing the number of men by 1 until you've exhausted all possibilities.

One more thing. Do I then sum 1 women and 4 men + ... 4 men & 1 women?

No. Look above.

 

[diracdelta]

Hmm, I tought something wrong.
But i got it now.
Thhanks !

 

[haruspex]

Hmm, I tought something wrong.
But i got it now.
Thhanks !

There are only 4 men, right? So doesn't any selection of 5 people satisfy (d)?

 

[diracdelta]

There are only 4 men, right? So doesn't any selection of 5 people satisfy (d)?

No.
So basicly I have this:
Five slots, where one is determined in advance to be women.
Lets say its first slot.
We can choose that women in 7 ways, so is this ok

$$7[\binom{6}{1}\binom{4}{3} + \binom{6}{2}\binom{4}{2} + \binom{6}{3}\binom{4}{1}] +\binom{6}{5}$$

 

[haruspex]

No.
So basicly I have this:
Five slots, where one is determined in advance to be women.
Lets say its first slot.
We can choose that women in 7 ways, so is this ok

$$7[\binom{6}{1}\binom{4}{3} + \binom{6}{2}\binom{4}{2} + \binom{6}{3}\binom{4}{1}] +\binom{6}{5}$$

The slots are not distinct, and neither are the women. According to (d), the only constraint is that at least one woman is chosen. Since 5 people are to be chosen and only 4 men are available, it is inevitable that at least one woman will be chosen.