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Three friends agreed to meet at 8 P.M. in on of the Spanish restaurants in town. Unfortunately they forgot to specify the name of the restaurant. If there are 10 Spanish restaurants
(a) find the number of ways they could miss each other
(b) find the number of ways they could meet
(c) find the number of ways at least two of them meet each other.

I got 10 * 3 = 30 ways for (a) for (b), I got 10 ways. how do I do (c)?

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I suggest you start by considering a simpler problem of the same type, say with 3 restaurants and 3 friends. See if the same method produces the correct answer in this simpler case. – MJD May 31 '12 at 2:28
Does "miss" mean "not all three friends end up together" or "all three friends end up in separate restaurants"? – Austin Mohr May 31 '12 at 2:28
yes they all miss each other, as in they went to different restaurants – count May 31 '12 at 2:30
Figure out the complement of (c). – copper.hat May 31 '12 at 2:31
up vote 3 down vote accepted

I’m afraid that you’ll have to start by redoing (a) and (b): both are wrong. Call the friends $A,B$, and $C$.

(a) I’m assuming that what’s wanted here is the number of ways in which they completely miss one another: no two go to the same restaurant. There are $10$ restaurants that $A$ could choose. $B$ could then choose any of the remaining $9$, and $C$ could choose any of the $8$ not chosen by $A$ or $B$. Thus, there are $10\cdot9\cdot8=720$ different ways in which they could completely miss one another.

(b) In order for them to meet successfully, they must all choose the same restaurant. There are $10$ restaurants, so there are just $10$ ways in which they can meet successfully. Alternatively, we can apply the same kind of analysis as in (a): $A$ has a choice of $10$ restaurants, but after that $B$ and $C$ have only $1$ choice each, if the meeting is to take place, so the total number of ways is $10\cdot1\cdot1=10$.

(c) This is most easily done by considering the complementary (opposite) event: no two of them meet. In (a) we calculated the number of ways in which this can happen. If there are $n$ different ways in which could they choose restaurants, regardless of how many of them meet, the number of ways in which at least two of them meet must be $n-720$. What’s $n$?

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Is $n=1000$?.... – count May 31 '12 at 2:57
@count: Yes, it is, so the answer to (c) is $1000-720=280$. – Brian M. Scott May 31 '12 at 2:59

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