In a team of 9 employees, 5 are members of the union. If the

Question

In a team of 9 employees, 5 are members of the union. If the team members must form a committee of 4, and if at least 2 of the committee members must be members of the union, how many committees are possible?

Can someone help? Thanks.

ian7777 · Answer

I think it's 60.

It's a combinations questions, since the order of the committee doesn't matter. Then you have to see that it's a combo question where the committee actually comes from 2 groups: union and non union members. As a rule, when you've got a combos question from more than one group, you need to do each group separately, and then multiply the answers together.

There are 5 union members, and 4 non union members, and each group will have 2 spots on the committee.

I don't like to use any formulas for combinations. I like to just put down the number of spaces there are to fill up, and then divide the answer by the (number of spaces)!. That's the easiest way to do combo questions. (If it's not clear why, I can explain it in more detail if you'd like).

So in this one, you've got: 

_ _ _ _
U U N N 

Where each space represents a spot on the committee. Then fill the spaces with the right number: (5 x 4)/2! x (4 x 3)/2! = 5 x 2 x 2 x 3 = 60.

cs77 · Answer

Hi Ian,

I totally agree with your answer if the question was specifying for only the scenario with 2 spots each on the committee. However, the "at least" part changes it, so you must account for the case where there are 2 union members on the committee (leaving 2 non-union members), 3 union members (leaving 1 non-union member), and 4 union members (leaving 0 non-union members).

So I think you have to count the combinations you calculated below for 2 union members / 2 non-members, but you have to add all of the other scenarios as well... this is the part I wasn't sure about how to do.... do you just take the combinations for each of the individual scenarios, calculate them, and add them up?

ian7777 · Answer

DOH!!

I guess that's what happens when you wake up and do math first thing in the morning! You're right, it is at least, and there are two ways to do it. It's actually not that complicated to do it your way, since there's a bunch of cancellation anyway. So you'd get the 60 from 2 members, and 40 from 3 members, and 5 from 4 members, and add them to get 105. Alternatively (and I don't think I'd do it this way, but it is possible), because it's "at least" that means that you could take away anything that DOESN'T fit the scenario. So that means you wouldn't want possibilities with no union members, or with just one union member. In that case, you could figure out how many total committees there are all together ( /4! = 126) and then subtract out the number of committees with only non union members (1) and the number with only one union member (20) and you'd get 105, also. But considering the fact that if you're getting this stuff, the first way takes about 45 seconds all together, I think it's an appropriate approach to a GMAT problem.

cs77 · Answer

Ok, thks for the explanation. I see both ways and I got 105 combos as well.

pakoo · Answer

Is the ansewer 105???

total ways 9C4 = 126
Ways in which Less than 2 are selected
5C0*4c4 +5C1*4C3 = 21

Wasys in which "atleast 2"= 126-21=105

srijay007 · Answer

Total combinations - como only 1 from union - combo noone from union

9C4- 5C1*4C3 - 1 = 105 is my ANS

jpv · Answer

1st Approach:
When 2 union members are selected: 5C2 * 4C2 = 60
When 3 union members are selected: 5C3 * 4C1 = 40
When 4 union members are selected: 5C4 = 5

Total = 105

2nd Approach: Same as Srijay.

In a team of 9 employees, 5 are members of the union. If the

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