A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at

[BTGmoderatorDC]

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

[Jay@ManhattanReview]

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

• 1 senior + 2 juniors:

# of ways = 4C1*6C2 = 4*(6*5)/(1*2) = 60;

• 2 seniors + 1 junior:

# of ways = 4C2*6C1 = (4*3)/(1*2)*6 = 36;

• 3 seniors + no junior:

# of ways = 4C3 = 4C1 = 4

Total no. of ways/groups = 60 + 36 + 4 = 100.

The correct answer: B

Hope this helps!

-Jay
_________________
Manhattan Review

Locations: Manhattan Review Bangalore | GMAT Prep Chennai | GRE Prep Himayatnagar | Hyderabad GRE Coaching | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

[Scott@TargetTestPrep]

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