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 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: \(C^3_5*C^2_9=360\);
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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