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by [email protected] » Sun Oct 05, 2014 11:26 am
Hi DrewNYCGmat,

This is a layered Combination Formula question that requires you to do 3 separate calculations...

The Combination Formula = N!/[K!(N-K)!]

We have 4 Senior partners and 6 Junior partners. We're asked for the number of combinations of 3 partners with AT LEAST 1 Senior partner. Here are the 3 options that we need to calculate:

1) 1 Senior and 2 Juniors = (4c1)(6c2) = (4)(15) = 60

2) 2 Seniors and 1 Junior = (4c2)(6c1) = (6)(6) = 36

3) 3 Seniors = (4c3) = 4

Total options = 60 + 36 + 4 = 100

Final Answer: B

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by DrewNYCGmat » Sun Oct 05, 2014 11:33 am
Thanks Rich! Very helpful
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by GMATGuruNY » Sun Oct 05, 2014 11:59 am
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
Alternate approach:

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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