October 16th, 2017, 12:04 PM  #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, 02:06 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625 
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. 
October 16th, 2017, 02:20 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, 02:30 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,202 Thanks: 1157 
At least 1 man $\displaystyle \sum_{m=1}^5~\dbinom{16}{m}\dbinom{8}{5m}=42448$ Last edited by romsek; October 16th, 2017 at 02:32 PM. 
October 16th, 2017, 02:45 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, 02:53 PM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 2,202 Thanks: 1157  Quote:
Last edited by romsek; October 16th, 2017 at 03:03 PM.  
October 16th, 2017, 05:37 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, 05:48 PM  #8 
Senior Member Joined: Sep 2015 From: USA Posts: 2,202 Thanks: 1157 
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 $n_m=\dbinom{12}{m}\dbinom{18}{8m}$ since there can be anywhere from 3 to 8 men in the group we take the sum of these to obtain $n_{38} = \displaystyle \sum_{m=3}^8~\dbinom{12}{m}\dbinom{18}{8m}$ as I wrote in the previous post. to help calculating this the formula is $\dbinom{n}{k} = \dfrac{n!}{k!(nk)!}$ 
October 16th, 2017, 09:42 PM  #9 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 
I found that on the TI84 Plus I must press ALPHA > F2 to find sigma. But what about those other notations?

October 16th, 2017, 09:49 PM  #10 
Senior Member Joined: Sep 2015 From: USA Posts: 2,202 Thanks: 1157  This is how to do factorial on a ti84 The $\Sigma$ isn't something you'll enter into your calculator. It means a sum. You can read about sigma notation here 

Tags 
combinations 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How many different combinations possible?  moistsnailtrail  Algebra  1  November 5th, 2013 09:14 AM 
help with combinations please  TheThinker  Algebra  1  March 29th, 2013 11:30 AM 
Complex Combinations Within Combinations Problem?  bugrocket  Advanced Statistics  2  January 23rd, 2011 05:02 PM 
Combinations within combinations possibilities?  aimpro2000  Advanced Statistics  2  September 20th, 2010 03:12 PM 
Am i doing this right? (Combinations)  Maria88  Number Theory  0  December 31st, 1969 04:00 PM 