I have
$$ P(n,r) = \frac{n!}{(n-r)!}, \qquad C(n,r) = \frac{n!}{r!(n-r)!}, \qquad C(n+r-1,r) = \frac{(n+r-1)!}{r!(n-1)!}.$$
Okay, how to get understand those, because I confusing. Let me help me with an exercise because I can't get which one of those are.
There is a student which has 3 lessons A, B, C and he has 30 days total. For every lesson he can give 5 until 15 days (with not playing role which exactly are). With which can give his time?
I will not do the $P(n,r)$ because I think it should play role the days. I am not sure. Could you help me with it?
$P(n,r)$ is the number of ways to pick $r$ items out of $n$ while noting the order. Imagine filling in numbers $1$ through $r$ into some subset of $n$ boxes.
$C(n,r)$ is the number of ways to pick $r$ items out of $n$ without noting the order. Imagine ticking $r$ out of $n$ boxes.
$C(n+r-1,r)$ is the number of ways to partition the number $r$ into $n$ summands, noting the order of summands. You can also think of this as follows: you have $r$ “summand” boxes and $n-1$ “separator” boxes. Use the formula above to count the ways of arranging these boxes. Then count the number of boxes between the separators. Each run from one separator to the next forms one summand. And the run from the beginning to the first separator, and from the last separator to the end, form another two summands, for a total of $(n-1)+1=n$ distinct runs, i.e. $n$ summands.
So now you want to split the 30 days they have into categories $A,B,C$. That would be $C(32,30)$. But you want each category to be at least $5$ and at most $15$. So subtract $5$ from each and you are partitioning $30-3\times 5=15$ using $C(17,15)$. But then you still have scenarios where one of your categories has more than $10$. How many? Well, if $A$ has at least $11$, then there are $4$ more to distribute to $B,C$ so you have $C(6,4)$. Same for when $B$ or $C$ has at least $11$.
Total count is ways to have $5$ in each categroy, minus ways where the limit of $15$ is exceeded. Can you come up with the formula now?