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quantitative question about combination :s

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quantitative question about combination :s [#permalink] New post 14 Nov 2013, 12:04
Hi everyone, I already posted my question on one topic but i had no answer.. i would be really grateful if someone would make me understand.

here is the question : a-group-consists-of-5-officers-and-9-civilians-if-you-must-91779.html
"A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?"

and here is the answer :


To meet the conditions we can have only 2 cases:
2 officers and 3 civilians: C^2_5*C^3_9=840;
3 officers and 2 civilians: C^3_5*C^2_9=360;

Total: 840+360=1,200.

so 3600 +1200 = 4800

but the thing I don't get is why we don't divide C^2_5*C^3_9 and C^3_5*C^2_9 by 2! as it is explain in this question? six-highschool-boys-gather-at-the-gym-for-a-modified-game-161915.html

where it is explained : when we make a combination and we don't want to order, we divide by n!

please help :(
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Re: quantitative question about combination :s [#permalink] New post 14 Nov 2013, 20:08
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Briefly put, the difference between the problem posted here and the one you linked to is that we are not forming multiple committees here, whereas there we formed multiple teams.

The two possible types of committees here are alternatives to one another. You aren't forming committees of 2 officers and 3 civilians AND of 3 officers and 2 civilians, rather you are forming either a committee of 2 officers and 3 civilians OR of 3 officers and 2 civilians.

In the basketball problem you are forming three teams. Here, you are forming a single committee, so there is no selection at the final stage.
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Re: quantitative question about combination :s [#permalink] New post 14 Nov 2013, 21:03
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oss198 wrote:
Hi everyone, I already posted my question on one topic but i had no answer.. i would be really grateful if someone would make me understand.

here is the question : a-group-consists-of-5-officers-and-9-civilians-if-you-must-91779.html
"A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?"

and here is the answer :


To meet the conditions we can have only 2 cases:
2 officers and 3 civilians: C^2_5*C^3_9=840;
3 officers and 2 civilians: C^3_5*C^2_9=360;

Total: 840+360=1,200.

so 3600 +1200 = 4800

but the thing I don't get is why we don't divide C^2_5*C^3_9 and C^3_5*C^2_9 by 2! as it is explain in this question? six-highschool-boys-gather-at-the-gym-for-a-modified-game-161915.html

where it is explained : when we make a combination and we don't want to order, we divide by n!

please help :(



Question 1: "you must select a committee of 5 people"
Question 2: "Three teams of 2 people each will be created."

In question 1, you have to make one committee and you can select people for it in 2 different ways: You can have 3 officers, 2 civ or you can have 3 Civ, 2 Officers. You add these different ways to get the total number of ways in which you can make one committee.

In question 2, you need to make 3 teams. You select the first team in 6C2 ways, the second team in 4C2 ways and the third team in 2C2 ways. Note that by selecting the teams one after the other, we have numbered them as first, second and third. This we don't need since teams aren't distinct. That is why we divide by 3!.
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Re: quantitative question about combination :s [#permalink] New post 16 Nov 2013, 11:29
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Thank you so much to both of you to have taken the time to answer me :)
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Re: quantitative question about combination :s [#permalink] New post 18 Nov 2013, 05:49
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oss198 wrote:
Thank you so much to both of you to have taken the time to answer me :)


Happy to help. Keep the good questions coming!
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Re: quantitative question about combination :s   [#permalink] 18 Nov 2013, 05:49
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