BTGmoderatorDC wrote: ↑Mon Mar 02, 2020 6:09 pm
A firm has 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
OA
B
Source: GMAT Prep
Solution:
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
