number of ways to sit 3 children in 5 available seats? My math teacher put a problem on my review packet and she told us to cross it out because she couldn't figure it out herself. However, both of us knew it had a solution. The problem was as follows:
How many possible ways can you sit 3 children in 5 available seats? (There was more background information, but that is not needed to understand the problem).
I tried using the counting principle and did 5P3. The result came out as 60.
I showed that to the teacher and she said that there is probably more to it. Was my answer correct? If not, what is the solution and more importantly, how do you get to the correct solution?
 A: There are several ways to approach this sort of problem. The other answers point out two of them, although they don't offer much explanation. But I'll point out two more ways, because part of the power and beauty of combinatorics is that you can often count the same thing in many different ways, and the more arguments you get exposed to, the better. (In more difficult problems, two students often come around to the same answer by approaching the problem very differently.)
First method: Name the three kids $A$, $B$, and $C$, and name the two seats that aren't used $E_1$ and $E_2$ (for empty). There is almost a one-to-one correspondence between ways of seating the three kids in the five chairs and the $5!$ arrangements of the five letters $A$, $B$, $C$, $E_1$, and $E_2$, aka an "anagram" of five letters. 
I say "almost a one-to-one correspondence," because there is a hitch: the empty seats are indistinguishable. What that means is that if you get the arrangement $ABE_1CE_2$, for example, you really have the same assignment of kids-to-seats as $ABE_2CE_1$. In other words, we can exchange the $E$'s and still have the same assignment of kids-to-seats. So using the anagrams of 5 letters leads to *over*counting the kids-to-seats assignments by a factor of 2 (because we can swap $E_1$ and $E_2$ in the end and get the same assignment), and so the answer is $\frac{5!}{2}$. This is indeed equal to $5P3$ (which is $\frac{5!}{2!}$ by definition, but $2!=2$).
This method, in other words, turns the problem into a problem you probably already know to do, i.e. counting the number of arrangements of letters in a word that contains repeated letters: for example, MISSISSIPPI. (There are 11 letters, but 4 indistinguishable I's, four indistinguishable S's, and two indistinguishable P's, so there are $\frac{11!}{4!\cdot 4!\cdot2!}$ anagrams of MISSISSIPPI.)
It also illustrates a deep point that most (I worry about saying "all") combinatorics problems are isomorphic to problems about anagrams.
Second method: The number of assignments of kids-to-seats can be found by breaking the problem down into two stages: (1) pick which 3 of the 5 seats you're going to put the kids in, and then multiply by (2) the number of ways to assign the kids to those 3 seats. There are ${5\choose 3}=10$ ways of picking 3 seats from 5, and there are $3!=6$ ways to assign the kids to those 3 seats once you've picked them. So using the multiplication principle, there are $10\cdot 6=60$ ways to assign the kids to seats.
A: Order the people as 1, 2, 3. You're essentially choosing three of the five chairs, and then imposing an order on those three chairs corresponding to the three people. So yes, ${}_5P_3$ is correct.
A: 60 sounds good assuming that they must all be seated. 
I would approach it from basic principle, You can put the first in one of 5, the second in one of 4 and the third in 1 of 3 = 5.4.3 = 60.
