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A five-member committee is to be formed from a group of five

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Intern
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A five-member committee is to be formed from a group of five [#permalink] New post 20 Sep 2008, 17:03
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?
answer is 1,200
Could someone please assist in the methodology of answering this?
thanks
Intern
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Re: Help with a Combination Problem [#permalink] New post 20 Sep 2008, 17:39
We are forming committee of 5 people from 5 military and 9 civilians. Also, the committee must form of at least 2 military personels and 2 civilians.

Lets form a team,
One team is 2 military personels and 3 civilians.
And, another team is 3 military personnels and 2 civilians.

So total team:
5C2 * 9C3 + 9C2 * 5C3 = 10 * 84 + 10 * 36
= 840 + 360
= 1, 200
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Re: Help with a Combination Problem [#permalink] New post 20 Sep 2008, 20:28
kidboc1 wrote:
A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?
answer is 1,200
Could someone please assist in the methodology of answering this?
thanks



Remember this -> When you have AND you MULTIPLY and when you have OR, you ADD


CASE1 -> we choose 3 out of 5 military officials AND 2 out of 9 civilians => 5C3 * 9C2

OR

CASE2 -> we choose 2 out of 5 military officials AND 3 out of 9 civilians => 5C2 * 9C3

So here CASE1 OR CASE2 = > 5C3*9C2 + 5C2*9C3 = 1200

Hope that helps
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"You have to find it. No one else can find it for you." - Bjorn Borg

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Re: Help with a Combination Problem [#permalink] New post 22 Sep 2008, 13:47
I was trying to solve it this way:

We need 5 people commitee:

# of way 2 officer can be choosen = 5 x 4
# of ways 2 citizens can be choosen = 9 x 8
# of ways the 5th person can be choosen = 5+9 (total ppl) - 4 (ppl already choosed) = 10

there total number of ways = (5 * 4 * 9 * 8 * 10) / 5 * 4 * 3 * 2 * 1 = 120

Can somebody please tell .. where am I missing the logic?
Re: Help with a Combination Problem   [#permalink] 22 Sep 2008, 13:47
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A five-member committee is to be formed from a group of five

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