How Do You Solve These Delegation Selection Problems in Combinatorics?

In summary, there are 210 ways to choose a delegation consisting of 3 women and 2 men out of 7 women and 4 men. For a delegation with an equal number of men and women, there are 329 ways to choose any number of people. For a delegation of 5 people with at least 3 women, there are 329 ways to choose any number of people. Finally, for a delegation of 5 people with one predetermined woman, there are 7 ways to choose the woman and 6 ways to choose the remaining 4 people, giving a total of 42 ways to choose the delegation.
  • #1
diracdelta
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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 :)

[tex]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}[/tex]
 
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  • #2
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?
 
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  • #3
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?
 
  • #4
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?
 
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  • #5
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?
 
  • #6
diracdelta said:
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.
diracdelta said:
One more thing. Do I then sum 1 women and 4 men + ... 4 men & 1 women?
No. Look above.
 
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  • #7
Hmm, I tought something wrong.
But i got it now.
Thhanks !
 
  • #8
diracdelta said:
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)?
 
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  • #9
haruspex said:
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

[tex]7[\binom{6}{1}\binom{4}{3} + \binom{6}{2}\binom{4}{2} + \binom{6}{3}\binom{4}{1}] +\binom{6}{5}[/tex]
 
  • #10
diracdelta said:
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

[tex]7[\binom{6}{1}\binom{4}{3} + \binom{6}{2}\binom{4}{2} + \binom{6}{3}\binom{4}{1}] +\binom{6}{5}[/tex]
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.
 

Related to How Do You Solve These Delegation Selection Problems in Combinatorics?

1. What is the difference between permutations and combinations?

Permutations and combinations are both ways of counting the number of possible arrangements or selections of a given set of items. The main difference is that permutations take into account the order of the items, while combinations do not. In other words, with permutations, the order in which the items are arranged matters, while with combinations, only the selection of items matters.

2. How do I calculate the number of permutations?

The formula for calculating the number of permutations is n! / (n-r)!, where n is the total number of items and r is the number of items being arranged. This formula is used when order matters, such as when arranging a set of objects in a line or selecting a specific order of items from a larger set.

3. How do I calculate the number of combinations?

The formula for calculating the number of combinations is n! / (r!(n-r)!), where n is the total number of items and r is the number of items being selected. This formula is used when order does not matter, such as when selecting a group of people from a larger set or choosing a combination of items from a set without regard to the order in which they are selected.

4. Can permutations and combinations be used in real-world scenarios?

Yes, permutations and combinations are used in many real-world scenarios, such as in probability, genetics, and computer science. For example, calculating the number of possible outcomes in a game of chance, determining the number of possible genetic combinations for a trait, and creating unique passwords using a combination of letters and numbers.

5. Are there any shortcuts or tricks for solving permutation and combination problems?

Yes, there are various shortcuts and strategies for solving permutation and combination problems, such as using factorials, combinations with repetition, and the fundamental counting principle. Additionally, there are many online calculators and tools available to assist with more complex calculations.

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