Distributing 10 books to 5 schools The problem asks:
10 copies of a book are distributed in 5 schools.
1. In how many ways can these books be distributed among schools?
2. What if at least one book is assigned to each school?

What I'm thinking is "I have 10 books to distribute in 5 schools, no matter how, so perhaps I can even give all the books to a single school"
So my answer to the first question is $C'_{10,5}=C_{14,5}$.
For the second question I was thinking "I have to give at least one book at school (so $C_{10,5}$), the rest (5) to any school, no matter how, so $C'_{5,5}$ my final answer is $C_{10,5} C'_{5,5}$".
Actually,  I was wrong on both answers, the answers are: $$1)\ C_{14,10}\\2)\ C_{9,5}$$
Why?
 A: Denote the five schools by $1,2,3,4,5$ and let $x_1,\ldots,x_5$ be the number of books distributed to each school, respectively. Then we always have an equation: $$x_1+x_2+x_3+x_4+x_5=10.$$ Our goal is to find the number of non-negative integer solutions under some specific conditions.

*

*There is no restriction, so by the star-and-bars formula, the number of non-negative integer solutions is precisely $$\binom{10+5-1}{10}=\binom{14}{10}.$$


*Now we have a restriction that $x_i\geq 1$ for $i=1,2,3,4,5$, but the star-and-bars formula only works for the restriction $x_i\geq 0$. Therefore, we define $y_i:=x_i-1$ for each $i$. Then the system becomes $$y_1+y_2+y_3+y_4+y_5=5,$$ where $y_i\geq 0$ for all $i$. Clearly, there is a one-to-one correspondence between the solution to $y_i$'s and the solution to $x_i$'s. By the star-and-bars formula, the number of non-negative integer solutions is $$\binom{5+5-1}{5}=\binom{9}{5}.$$
For the star-and-bars formula, you can check the Wikipedia.
A: This is a simple combinatorics problem where, as there is no restriction on having at least 1 book per school or such we can just plug in this formula.
(n+r-1) Choose r , which in this case is 14 choose 5 which is 2002.
Now if you need at least 1 book per school, This means that we can control the distribution of 5 books in any order. So again buy the same formula we get 9 choose 5 which is 126.
This theorem can be proved by the stars and bars method ([Stars and bars method discription])1. Please note that if your training for olympiads this theorem is bound to come in handy.
