Problem Solving

Hi arvin825,

On Test Day, you would have 5 answers to choose from, which might help you to take an educated guess (or give you an idea of what the answer might be close to).

As is, this is a rare layered Combinatorics question. To answer it, you have to perform 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

[spoiler]4 + 36 + 60 = 100 different groups[/spoiler]

GMAT assassins aren't born, they're made,
Rich

arvin825 wrote:Can someone help with solving the following:

A 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? (2 groups are different if at least one member is different)

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

We are asked to find the number of groups with at least one senior partner.

"At least 1" means "one or more," so the group must have 1 or 2 or 3 senior partners.

Case 1: Exactly 1 senior partner

Recall that the group must have 3 partners. Therefore, in this case, we need to pick 1 senior partner from 4 senior partners and 2 junior partners from 6 junior partners. The number of ways this can be done is 4C1 x 6C2.

4C1 x 6C2 = 4 x (6x5)/2! = 4 x 15 = 60

Case 2: Exactly 2 senior partners
In this case, we need to pick 2 senior partners from 4 senior partners and 1 junior partner from 6 junior partners. The number of ways this can be done is 4C2 x 6C1.

4C2 x 6C1 = (4x3)/2! x 6 = 6 x 6 = 36

Case 3: Exactly 3 senior partners

In this case, we need to pick 3 senior partners from 4 senior partners and no junior partners from 6 junior partners. The number of ways this can be done is 4C3 x 6C0.

4C3 x 6C0 = (4x3x2)/3! x 1 = 4 x 1 = 4

Thus, the total number of ways to form a group in which there is at least 1 senior partner = 60 + 36 + 4 = 100.

Alternate Solution:

It must be true that:

The total number of ways to form a group of 3 partners = (The number of ways in which the group would have at least 1 senior partner) + (The number of ways in which the group would have no senior partners).

Therefore:

The number of ways in which the group would have at least 1 senior partner = (The total number of ways to form a group of 3 partners) - (The number of ways in which the group would have no senior partners).
If the group of 3 has all junior partners, and there are 6 junior partners total, then the group of all junior partners can be made in 6C3 ways.

6C3 = (6 x 5 x 4)/3! = 5 x 4 = 20

The total number of groups of 3 that can be formed from 10 partners is 10C3.

10C3 = (10 x 9 x 8)/3! = 5 x 3 x 8 = 120

Thus, the number of ways to form a group of 3 in which there is at least 1 senior partner = 120 - 20 = 100 ways.

Answer: B