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.

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)

Two methods explained already so here comes another perspective to look at it

With Exactly 1 Senior: Total ways of choosing = 4x6x5/(2!) = 60 [2! in denominator excludes arrangements of 2 Juniors]
With Exactly 2 Senior: Total ways of choosing = 4x3x6/(2!) = 36 [2! in denominator excludes arrangements of 2 Seniors]
With Exactly 3 Senior: Total ways of choosing = 4x3x2/(3!) = 4 [3! in denominator excludes arrangements of 3 Seniors]

Total favorable cases = 60+36+4 = 100

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