marcodonzelli wrote:

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

This is how I solved this question and got the answer incorrect.

Total Combinations -

(the number of combinations in which A and B are always together) = Number of ways in which A and B won't work together

Assuming that A and B are the two consultants who don't want to work together.

Number of ways to choose 1 PM out of 2 candidates x Number of ways to choose 1 TL out of 3 candidates x Number of ways to choose 4 Consultants out of 7 candidates = Total

2 ways x 3 ways x 35 ways(7!/4!.3!) = 210 ways.

I got this right. However, somewhere in the next steps is where I made a mistake.

Number of ways in which A and B will work together and the rest two positions can be filled with the remaining 5 candidates.

2 x 3 x

1 x 1 x 5 x 4 So I subtracted 120 from 210.

However after following the answers here, I understood that for the remaining two positions, instead of doing a 5C2, I chose to fill them in 5 ways and 4 ways.

I would like to understand: 1) what sort of an error am I committing. 2When should I follow the approach I used above and when should I do a 5C2 logic.

Suggestions welcome. Thanks!

Regards,

Arvind.

I had the same query, but the concept of ordered and unordered selections helped me clear that doubt.

when we choose to us 5 x 4 to select the remaining two, it implies that the order of selection is important, whereas, when we use the formula 5C2 it implies that the order of selection is unimportant.

Which is what brings the difference in the answer. For this particular problem, the order of selection is unimportant and therefore the formula 5C2 will yield the correct answer.