Counting techniques I am preparing for an exam and I came across this problem. I am a little confused.
Give the expression of ways to distribute 15 distinguishable balls into five
distinguishable boxes so that the boxes have one, two, three, four, and five balls in
them, respectively. Assume there is no restriction on which box gets one ball, which
box gets two balls, etc.
How do I approach such  a problem?
EDIT: Would the answer to this be $15!/(1!\cdot2!\cdot3!\cdot4!\cdot5!)$? I'm confused about the assumption provided at the end. 
 A: Choose the lucky person who will end up with $5$ balls. This can be done in $\binom{5}{1}$ ways. Then choose the balls the person will get. This can be done in $\binom{15}{5}$ ways.
Now choose the person who will get $4$ balls. This can be done in $\binom{4}{1}$ ways. Choose the balls the person will get. This can be done in $\binom{10}{4}$ ways. And so on. We end up with
$$\binom{5}{1}\binom{15}{5}\binom{4}{1}\binom{10}{4}\binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2}\binom{1}{1}\binom{1}{1}.$$
The last two entries are a bit superfluous!
Remark: A little simpler is to divide the balls into $5$ piles, with $5,4,3,2,1$ in the piles. We can write down the number of such divisions as a multinomial coefficient, or as $\binom{15}{5}\binom{10}{4}\binom{6}{3}\binom{3}{2}$. Then multiply by $5!$ for the ways to distribute the piles among the $5$ people. 
The answer you provided probably used the sort of reasoning described above. For if you express the binomial coefficients in terms of factorials and simplify, you will find that $\binom{15}{5}\binom{10}{4}\binom{6}{3}\binom{2}{1}=\frac{15!}{5!4!3!2!1!}$. This answer should  be multiplied by $5!$. 
