My Math Forum Problem on Permutation Combination

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 July 4th, 2011, 12:50 PM #1 Newbie   Joined: Jul 2011 Posts: 2 Thanks: 0 Problem on Permutation Combination From a group of 4 Men and 4 Women a committee of 5 members is to be formed. Find the no. of ways of selecting the committee with at least 3 women such that at least 1 women holds the post of either a president or a vice - president ? It is known that the 5 member committee will have 1 president , 1 vice - president and 3 secretaries.
 July 4th, 2011, 11:15 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Problem on Permutation Combination You can either pick (a) four women, i.e. all of them, or (b) three women. a) Choose the man, and you have five people from which to pick the president and then the vice-president. b) Choose the woman who is left out, and two men. Then pick ...(i) a female president and female vice-president, or ...(ii) a female president and male vice-president, or ...(iii) a male president and female vice-president. Try this and show your results (with full working).
July 5th, 2011, 03:29 AM   #3
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Re: Problem on Permutation Combination

Hello, raminder!

Quote:
 From a group of 4 men and 4 women a committee of 5 members is to be formed. Find the number of ways of selecting the committee with at least 3 women such that at least 1 women holds the post of either president or vice-president? It is known that the 5-member committee will have 1 president, 1 vice-president and 3 secretaries.

I will refer to the offices of President, Vice President and Secretary as "P", "VP" and "S", respectively.
I will assume that the three secretarial positions are identical (interchangeable).

We want at least 3 women on the committee.
There are two cases.

$\text{(1) 4 women, 1 man}$
[color=beige]. . [/color]$\text{There are: }\:{4\choose4}{4\choose1} \:=\:1\,\cdot\,4\:=\:4\text{ choices.}$
[color=beige]. . [/color]$\text{They can be assigned offices in: }\:5\,\cdot\,4\,\cdot\,1\:=\:20\text{ ways.}$[color=beige] .[/color][color=red]**[/color]
$\text{Hence, there are: }\,4\,\cdot\,20 \:=\:80\text{ committees with 4 women and 1 man.}$

$\text{(2) 3 women, 2 men}$
[color=beige]. . [/color]$\text{There are: }\:{4\choose3}{4\choose2} \:=\:4\,\cdot\,6\:=\:24\text{ choices.}$
[color=beige]. . [/color]$\text{They can be assigned offices in: }\:5\,\cdot\,4\,\cdot\,1 \:=\:20\text{ ways.}$
$\text{So there are: }\,24\,\cdot\,20 \:=\:480\text{ committees with 3 women and 2 men.}$

$\text{But among these committees are some that we }don't\text{ want.}$
$\text{Namely, those in which the two men occupy the positions of P and VP.}$
[color=beige]. . [/color]$\text{P can be filled by any of the 4 men: }\:4\text{ choices.}$
[color=beige]. . [/color]$\text{VP can be filled by any of the other 3 men: }\:3\text{ choices.}$
[color=beige]. . [/color]$\text{The 3 S's can filled by any 3 of the 4 women: }\:{4\choose3} \,=\,4\text{ choices.}$
$\text{So there are: }\:4\,\cdot\,3\,\cdot\,4\:=\:48\text{ committees that we }don#39;t\text{ want.}$

$\text{Hence, there are: }\:480\,-\,48 \:=\:432\text{ committees that we }do\text{ want.}$

$\text{Therefore: }\:80\,+\,432 \:=\:512\text{ committees with at least 3 women}$
[color=beige]. . [/color]$\text{in which at least one woman holds the post of P or VP.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[color=red]**[/color]

Once we have selected the five people on the committee,
we can assign the posts to them like this:
[color=beige]. . [/color]There are 5 choices for the President.
[color=beige]. . [/color]There are 4 choices for the Vice President.
[color=beige]. . [/color]The other 3 are assigned as Secretaries: 1 way.
Hence, there are:[color=beige] .[/color]$5\,\cdot\,4\,\cdot\,1\:=\:20\text{ assignments.}$

 July 5th, 2011, 10:44 AM #4 Newbie   Joined: Jul 2011 Posts: 2 Thanks: 0 Re: Problem on Permutation Combination To aswoods thanks got some clarity on the problem. To soroban thanks for the solution but still i need clarification on points you have made first for the committees that we do not want in 2nd case in which we have 2 men occupying the position of P and VP how can P be filled by 4 men and have 4 choices ? it can have only 2 choices right , because we only have 2 men in the committee now in this case and similarly for VP we should have only 1 choice because of the 2 men 1 has been selected for P so how come are you taking 3 choices for them ? similarly for S there should be only 1 choice 3 women 3 posts. so the figure to subtract must be 2.1.1 that is 2 so 480 -2 = 478 and total should be 478+ 80 = 558. please rectify me where am i going wrong ? Thanks in advance.

### what will be the number of ways of selecting the committee with atleast 3 women such that at least one women holds the post of either president or vice president

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