Balls and urns when urns are indistinguishable 
In how many ways can $100$ identical books be distributed into $10$ indistinguishable bags so that no two bags contain the same no. of books and no bag is empty?

My approach:
Though this looks like a simple balls and urns problem, it is not so. I would have easily been able to do the problem if it had not used the word "indistinguishable".
We have the equation $$\sum_{i=1}^{10} x_i=100$$ where $x_i\ge1$
So we can tackle this by replacing $x_i$ with $x_i^{'}+1$, to get $$\sum_{i=1}^{10} x_i^{'}=90$$ and then use $\binom{n+r-1}{r}$
But what to do with indistinguishable?
Edit:
So as "some SE user" told me to divide it by $10!$, I think the ans is $\frac{\binom{10+90-1}{90}}{10!}$
 A: Since the 10 bags are indistinguishable, the 100 books are identical, we may assume that the bags are arranged with respect to the their number of balls. Let $0< x_1<x_2<\cdots <x_{10}$ be the number of books in each bag. Then
$$\sum_{i=1}^{10}x_i=100$$
that is, by letting $x_i=y_i+i-1$,  $1\leq y_1\leq y_2\leq \cdots \leq y_{10}$
$$\sum_{i=1}^{10}y_i=100-\sum_{i=1}^{10}(i-1)=55.$$
So you should count the number of integer partitions of 55 in exactly 10 terms, that is $P(55,10)$ which satisfies the recurrence
$$P(n,k) = P(n-1,k-1) + P(n-k,k)$$
see Integer partition of n into k parts recurrence
P.S. Your problem is asking for the number of partitions of $n=100$ into $k=10$ distinct parts. The generating function of such numbers $a(n,k)$ is $$f_k(x)=\sum_{n\ge 0}a(n,k)x^n=\frac{x^{k+\binom{k}2}}{(1-x)(1-x^2)\dots(1-x^k)}.$$
According to WA, $a(100,10)=33401$.
A: In effect, you are counting the sequences of integers $(x_1,\ldots,x_{10})$
with $1\le x_1<x_2<\cdots<x_{10}$ and $x_1+\cdots+x_{10}=100$. If we set $y_k=x_k-k$, this is the same as the number of sequences of nonnnegative
integers $(y_1,\ldots,y_{10})$ with $y_1\le y_2\le\cdots\le y_{10}$
and $y_1+\cdots+y_{10}=45$. This is the number of partitions of $45$
into $\le10$ parts, or the number of partitions of $45$ into parts of
size $\le10$. It is the coefficient of $q^{45}$ in $\prod_{k=1}^{10}(1-q^k)^{-1}$.
