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How many different groups can be formed?

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by CappyAA » Tue Aug 12, 2008 5:16 pm
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different.)

A. 48
B. 100
C. 120
D. 288
E. 600


Thanks,
Cappy
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by egybs » Tue Aug 12, 2008 5:53 pm
What's the total number of groups possible?

10C3 (because there are ten total people): 10!/(7!*3!) = 120


But that includes all the cases where there are no senior partners...

So let's figure out the number of cases where there are no senior partners....

6C3 (because there are 6 junior partners) = 6!/(3!*3!) = 20

So 120 - 20 = 100 or the number of possible groups where there is at least 1 senior partner.
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by gracezz » Wed May 12, 2010 7:04 pm
i went for this caculation

4!*9!/(7!*2!)=144 at least have one senior, right? so 4!, and you pick two out of the rest nine persons.

would you guys tell me at what point I am wrong with the puzzle?
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by chocoswiss » Sun Oct 03, 2010 2:56 am
Hi

Sorry I still dont understand Why i cant do as below:
1S2J: 4*6*5
2S1J: 4*3*6
3S: 4*3*2


I somehow always end up reverting to the slot method
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by GMATGuruNY » Sun Oct 03, 2010 3:11 am
chocoswiss wrote:Hi

Sorry I still dont understand Why i cant do as below:
1S2J: 4*6*5
2S1J: 4*3*6
3S: 4*3*2

I somehow always end up reverting to the slot method
Since order doesn't matter, you need to do some dividing in order to account for the duplicate combinations:

1S2J = 4 * (6*5)/(1*2) = 60
2S1J = (4*3)/(1*2) * 6 = 36
3S = (4*3*2)/(1*2*3) = 4

60+36+4 = 100.

You also could use the slot method this way:

Total possible groups of 3 = (10*9*8)/(1*2*3) = 120
Total possible groups of 3 with only junior partners = (6*5*4)/(1*2*3) = 20
Total possible groups of 3 with at least 1 senior partner = 120-20 = 100.
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