Ways to select three-man teams

In a competition there are 18 competitors. Answer the following:

A) During the first day they're competing in three-man teams (total of 6 teams). How many ways are there to select the teams?

B) If the main sponsor of the event demands that the four best ranked players mustn't play in the same groups, how many ways can you then select the teams?

I've tried the following:

A) There are (18*17*16 + 15*14*13 + 12*11*10 +...+ 3*2*1) ways to select the teams, and if the order of the groups mattered we would just multiply that wit 6!

However,this gives me approximately 700,000 ways, whereas the correct number of ways according to the book is ~190,000,000

B) The four best players must each be in a different group. Therefore we "lock" their positions. (1*14*13 + 1*12*11 + 1*10*9 + 1*8*7 + 6*5*4 + 3*2*1) gives us the ways to select the teams, and if the order of the groups matters, we multiply that with 6!

Once again, this value is far too small according the the book. Either my calculations are way off or the book is wrong.

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Homework? What have you tried? – Tim Seguine Feb 13 '14 at 21:04
@TimSeguine Nope, not homework. My teacher and I couldn't get our heads around this today and I was simply curious to see how others would approach the problem. – Mylleranton Feb 13 '14 at 21:06
You are leaving out the fact that the order of the players inside of each team doesn't matter. A team with Bob, Alice and Jane is the same as a team with Jane, Bob and Alice. – Tim Seguine Feb 13 '14 at 21:13
So I would take 18 over 3 plus 15 over 3....until 3 over 3? That would give me an even less value. – Mylleranton Feb 13 '14 at 21:22
You are "zigging" where you should be "zagging", see my answer. – Tim Seguine Feb 13 '14 at 21:32

I will use the binomial coefficient notation. $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is read "n choose k", the number of ways to choose $k$ items without replacement from a pool of $n$ distinct objects.

A) There are $\binom{18}{3}$ ways to choose the first team, $\binom{15}{3}$ ways to choose the second, etc. this gives $$\binom{18}{3}\binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$ ways to choose 6 teams. But the order of the teams does not matter, so we have to divide by the number of permutations of the teams ($6!$). This gives an end result of 190,590,400, which appears to be the answer your book is expecting.

Basically, you are adding where you should be multiplying. See dmk's answer for a little more about that.

See if you can get started yourself on the second part. If you need more help, let me know.

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Thanks a lot! Figured out B) as well with this reasoning. Guess I got to dug in on the problem I lost my bearings. – Mylleranton Feb 13 '14 at 21:44
@Mylleranton Always glad to help. – Tim Seguine Feb 13 '14 at 21:47

For (a), as an earlier comment stated, you're making order matter by using a permutation. So, for example, you have $18$ choices for the first person on Team 1, $17$ choices for the second, and $16$ choices for the third. Just as you wrote, this makes $18\cdot 17\cdot 16$ choices. However, from those three people, we could have chosen the first person $3$ ways, the second $2$ ways, and the final person $1$ way. Therefore, $18\cdot 17\cdot 16$ overcounts the number of teams by a factor of $3\cdot 2 \cdot 1$. That means there are $\frac{18\cdot 17\cdot 16 }{3\cdot 2 \cdot 1}$ ways to choose the first team. This is equivalent to the binomial coefficient $\binom{18}{3}$, if you're familiar with combinations.

However, following this idea for each team will make your answer smaller, whereas your answer is too small to begin with. What you want to do is multiply those fractions rather than add them. Here we're determining one team and then determining another and then determining a third... Often, when an "and" appears as part of how you would complete a task (as above), that means you multiply. An "or", on the other hand, tells you to add; for example, if you wanted to make only one team, and you wanted to know how many ways there were to create a team of either three or four people from 18, you'd calculate

$$\frac{18\cdot 17\cdot 16 }{3\cdot 2 \cdot 1} + \frac{18\cdot 17\cdot 16\cdot 15}{4\cdot 3 \cdot 2 \cdot 1}$$

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This has some quite generally applicable reasoning. Thanks a lot! – Mylleranton Feb 13 '14 at 21:47
+1 for explaining the "and/or" thing, which I just sort of glossed over in my answer. – Tim Seguine Feb 13 '14 at 21:50