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 October 16th, 2017, 11:04 AM #1 Senior Member   Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 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?
 October 16th, 2017, 01:06 PM #2 Global Moderator   Joined: May 2007 Posts: 6,524 Thanks: 587 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. Thanks from topsquark
 October 16th, 2017, 01:20 PM #3 Senior Member   Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 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.
 October 16th, 2017, 01:30 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 1,974 Thanks: 1025 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 01:32 PM.
 October 16th, 2017, 01:45 PM #5 Senior Member   Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 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
October 16th, 2017, 01:53 PM   #6
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Quote:
 Originally Posted by Opposite 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 02:03 PM.

 October 16th, 2017, 04:37 PM #7 Senior Member   Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 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.
 October 16th, 2017, 04:48 PM #8 Senior Member     Joined: Sep 2015 From: USA Posts: 1,974 Thanks: 1025 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 $\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}$
 October 16th, 2017, 08:42 PM #9 Senior Member   Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 I found that on the TI-84 Plus I must press ALPHA -> F2 to find sigma. But what about those other notations?
 October 16th, 2017, 08:49 PM #10 Senior Member     Joined: Sep 2015 From: USA Posts: 1,974 Thanks: 1025 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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