An engagement team consists of a project manager, team

BTGmoderatorLU wrote:Source: Manhattan Prep

An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

A. 25
B. 35
C. 150
D. 210
E. 300

The OA is C.

For this question, let's first ignore the restriction regarding the two consultants who refuse to work together and find the total number of different teams.
Then we'll determine how many of those teams break the rule about the two consultants.

In other words, # of "good" teams = Total number of teams (ignoring the restriction) - # of "bad" teams (that break the rule)

Total number of teams (ignoring the restriction)
Take the task of building teams and break it into stages.

Stage 1: Select a project manager
There are 2 candidates for this position, so we can complete stage 1 in 2 ways

Stage 2: Select a team leader
There are 3 candidates for this position, so we can complete stage 2 in 3 ways

Stage 3: Select the four consultants
Since the order in which we select the consultants does not matter, we can use combinations.
We can select 4 consultants from 7 consultants in 7C4 ways (35 ways)

If anyone is interested, we have a video on calculating combinations (like 7C4) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus build our team) in (2)(3)(35) ways ( = 210 ways)


# of "bad" teams (that break the rule)
Take the task of building "bad" teams and break it into stages.

Stage 1: Select a project manager
There are 2 candidates for this position, so we can complete stage 1 in 2 ways

Stage 2: Select a team leader
There are 3 candidates for this position, so we can complete stage 2 in 3 ways

Stage 3: Select four consultants
Here, we want to break the rule. So, let's place the two bickering consultants on the team.
So, for this stage, we need only select two other consultants to join them.
Since the order in which we select the 2 remaining consultants does not matter, we can use combinations.
We can select 2 consultants from the remaining 5 consultants in 5C2 ways (10 ways)

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus build our "bad" team) in (2)(3)(10) ways ( = 60 ways)


So, # of "good" teams = 210 - 60
= 150
= C

Cheers,
Brent

BTGmoderatorLU wrote:Source: Manhattan Prep

An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

A. 25
B. 35
C. 150
D. 210
E. 300

We can select the project manager in 2 ways and the team leader in 3 ways. For the consultants, there are 7 candidates for 4 positions, but 2 of the 7 cannot be together.

We can use the following equation:

# of ways to select the consultants = # of ways with the 2 together + # of ways with the 2 not together.

The total number of ways to select the consultants is 7C4 = 7!/[4!(7-4)!] = 7!/(4!3!) = (7 x 6 x 5 x 4)/4! = 7 x 6 x 5 x 4)/(4 x 3 x 2) = 35.

The number of ways to select the consultants when the 2 are together can be found by choosing 2 people for the remaining slots from the 5 remaining candidates, which is given by 5C2 = (5 x 4)/2 = 10.

Thus, the number of ways to select the consultants when the 2 are not together is 35 - 10 = 25.

So, the total number of ways to select the teams is 2 x 3 x 25 = 150.

Answer: C