Guys, I need help with this question. I just can't figure out how the answer can be 24.

15. The Acme Plumbing Company will send a team of 3 plumbers to work on a certain job. The company has 4 experienced plumbers and 4 trainees. If a team consists of 1 experienced plumber and 2 trainees, how many different such teams are possible?

"If a team consists of 1 experienced plumber and 2 trainees, how many different such teams are possible?"

So, that would mean 4 plumbers x 4 trainees x 3 trainees (1 picked already), which would mean 48. Now, how is it 24 and not 48?

Since order does not matter, because one plumber is the same as another and one trainee is the same as another trainee, you first

Find all possible combinations of 2 trainees out of 4 possible , so in calc
4 nCr 2 = 6
Then you need one out of 4 plumbers at a time, so there are 4 possibilities
4 nCr 1 = 4

15. The Acme Plumbing Company will send a team of 3 plumbers to work on a certain job. The company has 4 experienced plumbers and 4 trainees. If a team consists of 1 experienced plumber and 2 trainees, how many different such teams are possible?

Best way to approach this:

Fact 1: There are 4 different experienced plumbers.
Fact 2: There are 6 different combinations of 2 trainees in a total group of 4 trainees. If this isn't obvious, let the 4 trainees be A,B,C,and D.

All the possible trainee pairs:

AB,AC,AD
BC,BD
CD

and each of these trainee pairs can go with any of the 4 experienced plumbers, so we multiply 6 X 4 for the total number of possible teams, which gives 24. A common error here would be to erroneously consider AB a different pair from BA, when really order doesn't matter in this problem. That is how you would get a wrong answer of 48.

## Replies to: Blue Book: Math, pg. 657, #15

You have to use combination... 4C1 x 4C2 = 4 x 4x3/2 = 24

In your method, exp + trainee + trainee and trainee + trainee + exp count as different teams, but they're the same.

Just to make sure though, how would one calculate this if the order mattered?

So 24 teams are possible.... then you do permutation for 3?

24 x 3! which equals 144. Can anyone verify this?

Find all possible combinations of 2 trainees out of 4 possible , so in calc

4 nCr 2 = 6

Then you need one out of 4 plumbers at a time, so there are 4 possibilities

4 nCr 1 = 4

6 x 4 = 24

Best way to approach this:

Fact 1: There are 4 different experienced plumbers.

Fact 2: There are 6 different combinations of 2 trainees in a total group of 4 trainees. If this isn't obvious, let the 4 trainees be A,B,C,and D.

All the possible trainee pairs:

AB,AC,AD

BC,BD

CD

and each of these trainee pairs can go with any of the 4 experienced plumbers, so we multiply 6 X 4 for the total number of possible teams, which gives 24. A common error here would be to erroneously consider AB a different pair from BA, when really order doesn't matter in this problem. That is how you would get a wrong answer of 48.