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April 30th, 2014, 05:44 PM  #1 
Newbie Joined: Apr 2014 From: USA Posts: 6 Thanks: 0  Committee of 5 members is to be formed
Committee of 5 members to be formed from 5 married couples. Find the # of committees when: 1) the selections random 2) there are more men than women 3) a particular couple is on the committee 
April 30th, 2014, 06:28 PM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, LuLu99! Quote:
$\displaystyle \quad {10\choose5} \:=\:\frac{10!}{5!\,5!} \:=\:252$ committees. Quote:
$\displaystyle \quad \begin{array}{cccc} \text{3M, 2W} & \text{4M, 1W} & \text{5M, 0W} \\ {5\choose3}{5\choose2} & {5\choose4}{5\choose1} & {5\choose5}{5\choose 0} \\ 10\cdot10 & 5\cdot 5 & 1\cdot 1 \end{array}$ Therefore: $\displaystyle \:100 + 25 + 1 \;=\;126$ committees. Quote:
If $\displaystyle A$ and $\displaystyle B$ are already on the committee, we must choose 3 more people from the remaining 8 people. There are: $\displaystyle \:{8\choose3} \,=\,56$ committees.  

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