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# quantitative question about combination :s

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Manager
Joined: 18 Jul 2013
Posts: 69

Kudos [?]: 97 [0], given: 114

Location: Italy
GMAT 1: 600 Q42 V31
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14 Nov 2013, 13:04
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!

Kudos [?]: 97 [0], given: 114

Economist GMAT Tutor Instructor
Joined: 01 Oct 2013
Posts: 69

Kudos [?]: 48 [1], given: 7

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14 Nov 2013, 21:08
1
KUDOS
Expert's post
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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Kudos [?]: 48 [1], given: 7

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14 Nov 2013, 22:03
1
KUDOS
Expert's post
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!

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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Kudos [?]: 17874 [1], given: 235

Manager
Joined: 18 Jul 2013
Posts: 69

Kudos [?]: 97 [1], given: 114

Location: Italy
GMAT 1: 600 Q42 V31
GMAT 2: 700 Q48 V38

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16 Nov 2013, 12:29
1
KUDOS
Thank you so much to both of you to have taken the time to answer me

Kudos [?]: 97 [1], given: 114

Economist GMAT Tutor Instructor
Joined: 01 Oct 2013
Posts: 69

Kudos [?]: 48 [0], given: 7

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18 Nov 2013, 06:49
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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