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# A certain law firm consists of 4 senior partners and 6

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Senior Manager
Joined: 08 Jun 2004
Posts: 498
Location: Europe
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A certain law firm consists of 4 senior partners and 6 [#permalink]  14 May 2006, 04:42
00:00

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
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 number 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
Director
Joined: 16 Aug 2005
Posts: 946
Location: France
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I got this question in my GMATPrep exam today and got it right

B
Senior Manager
Joined: 08 Jun 2004
Posts: 498
Location: Europe
Followers: 1

Kudos [?]: 20 [0], given: 0

Hey Prof is my approach correct? Thank you.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
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Kudos [?]: 186 [0], given: 0

Combis:

3 senior partner ---> # of ways = 4C3 = 4
2 senior partner + 1 junior partner ---> # of ways = 4C2*6C1 = 6*6 = 36
1 senior partner + 2 junior partner ---> # of ways = 4C1*6C2 = 4*15 = 60

Total = 4+36+60 = 100 groups
GMAT Club Legend
Joined: 07 Jul 2004
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Location: Singapore
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M8 wrote:
Guys what about 10C3 - 4C1*6C1?

What is your basis for picking 4C1 and 6C1. You're just picking 1 senior partner and 1 junior partner, so you'll still be one man short.

The correct method should be 10C3 - 6C3 = 120-20 = 100. We use 6C3 because 3 junior partners in the group is an illegal condition.
VP
Joined: 29 Dec 2005
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M8 wrote:
Guys what about 10C3 - 4C1*6C1?

=10C3 - 4C1*6C1 = 120 - 24 = 96 is not correct because 4C1*6C1 means you are selecting each senoir and junior from 4 seniors and 6 juniors. why should we choose 2 (s+j) from 10?

you can do as deowl did.

=10C3 - 6C3
=120-20
=100
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