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# of different groups of parters

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by GMATGuruNY » Sun Apr 26, 2015 2:53 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
Groups with AT LEAST 1 senior partner = all possible groups - groups with NO senior partners.

All possible groups:
From 10 partners, the number of ways to choose a group of 3 = 10C3 = (10*9*8)/(3*2*1) = 120.

Groups with no senior partners:
From 6 junior partners, the number of ways to choose a group of 3 = 6C3 = (6*5*4)/(3*2*1) = 20.

Thus:
Groups with at least 1 senior partner = 120-20 = 100.

The correct answer is B.
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by [email protected] » Sun Apr 26, 2015 5:04 pm
Hi binaras,

This is a rare layered Combinatorics question. The most efficient way to answer it is to use subtract the options that you "don't want" from the total possible options (the approach that Mitch used); you could also calculate what you DO want, but that requires 3 separate multi-step calculations...

We're given 4 senior partners and 6 junior partners. We're asked for the number of different groups of 3 (the clue that we'll need the Combination Formula) with one stipulation - there must be AT LEAST 1 senior partner. Here are the 3 calculations:

1) 3 senior partners = 4c3 = 4!/[3!1!] = 4 different groups

2) 2 seniors and 1 junior = (4c2)(6c1) = (4!/[2!2!])(6!/[1!5!) = (6)(6) = 36 different groups

3) 1 senior and 2 juniors = (4c1)(6c2) = (4!/[3!1!])(6!/[2!4!]) = (4)(15) = 60 different groups

4 + 36 + 60 = 100 different groups

Final Answer: B

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