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wrong.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?My friend Debbie Stier got it right -- ! After she explained it to me, C. and I saw what we were doing wrong (double-counting the trainee pairs, basically), too, but we wanted to watch the Khan video to see how he did it.
Turns out Salman Khan did the problem the same way C. and I did.
Which makes me feel somewhat better.
Best explanation for novices on College Confidential:
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.
5 comments:
Choose 1 from 4 then choose 2 from 4.
4c1 * 4c2 = 4*6 = 24
The problem is that I don't 'see' that -----
Once someone tells me to treat the two cases separately, fine.
Thanks!
At every step, you need to ask: does order matter?
If people are sitting in a line, order matters. If people are assigned to a bucket, or a room, or some other thing that's not ordered, then their order doesn't matter.
When order matters, then
A B
is different from
B A.
When order doesn't matter, those two cases are the same.
"4 choose 2" is a short hand for saying "of the 4 items, give me all the ways to choose 2 when THEIR ORDER DOES NOT MATTER."
I'm going to send a thank-you to Art of Problem Solving.
After having worked my way through nearly half their Introduction to Counting and Probability, this problem now seems simple.
I think I'll go see if the Gruber counting problems have gotten any easier.
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