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a 5 member committe is to be formed from a group of 5

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a 5 member committe is to be formed from a group of 5 [#permalink] New post 19 Aug 2008, 07:45
a 5 member committe is to be formed from a group of 5 military officers and 9 civilians. if the committee must include at least 2 civilians and 2 military officers, in how many different ways can the committee be chosen?

119

1200

3240

3600

14,400
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Re: combinations [#permalink] New post 19 Aug 2008, 07:53
pattydread wrote:
a 5 member committe is to be formed from a group of 5 military officers and 9 civilians. if the committee must include at least 2 civilians and 2 military officers, in how many different ways can the committee be chosen?

119

1200

3240

3600

14,400


Two combos = (3 civlians , 2 millitary) + ( 2 civilians, 3 military) = 9C3 * 5C2 + 9C2 * 5C3 = 84*10 + 36*10 = 1200
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Re: combinations [#permalink] New post 19 Aug 2008, 12:37
There are two combinations here in a 5 member committee.
2 civilians 3 military officers and 3 civilians and 2 military officers.
For the case I- the possible number of combinations 5C2*9C3=10*84=840
For case II- the possible number of combinations 5C3*9C2=10*36=360
Total number of combinations 840+360=1200
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Re: combinations [#permalink] New post 19 Aug 2008, 12:43
pattydread wrote:
a 5 member committe is to be formed from a group of 5 military officers and 9 civilians. if the committee must include at least 2 civilians and 2 military officers, in how many different ways can the committee be chosen?

119

1200

3240

3600

14,400


isnt it
First choose 2/9 civilians * choose 2/5 officers * the remaining
= 9C2 * 5C2 * (7+3)
= (9 * 8)/2 * (5*4)/2 * 10
= 3600
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Re: combinations [#permalink] New post 19 Aug 2008, 13:11
pattydread wrote:
a 5 member committe is to be formed from a group of 5 military officers and 9 civilians. if the committee must include at least 2 civilians and 2 military officers, in how many different ways can the committee be chosen?

119

1200

3240

3600

14,400


D

(5 choose 2)*(9 choose 2)*10 = 3600
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Re: combinations [#permalink] New post 19 Aug 2008, 13:57
sset009 wrote:

isnt it
First choose 2/9 civilians * choose 2/5 officers * the remaining
= 9C2 * 5C2 * (7+3)
= (9 * 8)/2 * (5*4)/2 * 10
= 3600


No. In your approach you are triple couting all the options. So you need to divide by 3 to get the correct answer.

Let us say m1-m5 are the 5 military guys and c1-c9 are the 9 civvis. In your expression,
m1 m2 + c1 c2 + m3 group is counted as being different from m1 m3 + c1 c2 + m2 and m2 m3 + c1 c2 + m1[/. In actuality they are all same m1m2m3 and c1c2.
So for each case you are taking 3 cases. Divide by 3 adn you will get the right answer.
Re: combinations   [#permalink] 19 Aug 2008, 13:57
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a 5 member committe is to be formed from a group of 5

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