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October 16th, 2017, 12:04 PM   #1
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Combinations?

A group of 24 people, 16 of which are men, is forming a 5 person committee. How many different ways can the committee be formed if at least 1 of the members must be men?

I thought the answer was, "(16 nCr 1) * (22 nCr 4), but it's not. What is the right equation?
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October 16th, 2017, 02:06 PM   #2
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There are a total of $\displaystyle \binom {24}{5} $ possible committees.
There are a total of $\displaystyle \binom {8}{5} $ possible committees where there are no men.
Subtract to get your answer.
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October 16th, 2017, 02:20 PM   #3
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24 nCr 5 = 42504.

42504 - (8 nCr 5) = 42448

That doesn't seem right. Applying this method to other problems of the same kind is telling me it's wrong.
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October 16th, 2017, 02:30 PM   #4
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At least 1 man

$\displaystyle \sum_{m=1}^5~\dbinom{16}{m}\dbinom{8}{5-m}=42448$

Last edited by romsek; October 16th, 2017 at 02:32 PM.
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October 16th, 2017, 02:45 PM   #5
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Here's a practice one. Maybe this will help more.

A group of 30 people, 12 of which are men, is forming a 8 person committee. How many different ways can the committee be formed if at least 3 of the members must be men?

The answer is 4202055, but I don't know how to get there.

30 nCr 8 = 5852925
5852925 - ( 18 nCr 8 ) = 5809167
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October 16th, 2017, 02:53 PM   #6
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Quote:
Originally Posted by Opposite View Post
Here's a practice one. Maybe this will help more.

A group of 30 people, 12 of which are men, is forming a 8 person committee. How many different ways can the committee be formed if at least 3 of the members must be men?

The answer is 4202055, but I don't know how to get there.

30 nCr 8 = 5852925
5852925 - ( 18 nCr 8 ) = 5809167
$\displaystyle \sum_{m=3}^8~\dbinom{12}{m}\dbinom{18}{8-m} = 4202055$

Last edited by romsek; October 16th, 2017 at 03:03 PM.
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October 16th, 2017, 05:37 PM   #7
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I don't know how to put that into a calculator. And I don't understand how the equation works. It's not the way we are learning in class.
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October 16th, 2017, 05:48 PM   #8
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suppose you have to form groups of 8 that had 1 male only, out of 12 men and 18 women, 30 total.

there are $\dbinom{12}{1}$ ways of selecting that one male

and then there are $\dbinom{18}{7}$ ways of selecting the 7 females out of the 18.

The number of such groups is the product of these numbers

$n_1 = \dbinom{12}{1}\dbinom{18}{7}$

Similarly if you have to form groups of 8 with $m$ males only you would have

$n_m=\dbinom{12}{m}\dbinom{18}{8-m}$

since there can be anywhere from 3 to 8 men in the group we take the sum of these to obtain

$n_{3-8} = \displaystyle \sum_{m=3}^8~\dbinom{12}{m}\dbinom{18}{8-m}$

as I wrote in the previous post.

to help calculating this the formula is $\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}$
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October 16th, 2017, 09:42 PM   #9
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I found that on the TI-84 Plus I must press ALPHA -> F2 to find sigma. But what about those other notations?
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October 16th, 2017, 09:49 PM   #10
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This is how to do factorial on a ti-84


The $\Sigma$ isn't something you'll enter into your calculator. It means a sum.

You can read about sigma notation here
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