Intuition on why this problem doesn’t have an analytic solution I recently overheard some people trying to work out a solution for a problem informally stated as follows:
In how many ways can you put $n$ indistinguishable particles into $n$ boxes with the condition that the number of particles in each box is nonincreasing to the right?
I was told that this has no analytic solution and relies on asymptotic methods from partition theory or something like that... Now I don’t have much experience with combinatorics and none with partition theory but intuitively this seems to me (at first glance at least) like it should have an analytic solution. Just wondering if anyone could give some insight as to why it does not. Thanks!
 A: As noted in the comments, if the number of particles per box is required to be strictly decreasing to the right, then the problem is trivial. If the number of particles is required to be nonincreasing to the right, then the number of ways to fill the boxes is precisely the number of partitions of $n$:
Let $m_k$ denote the number of particles in box $k$. Then putting the particles in the boxes with a nonincreasing way is the same as writing
$$n=\sum_{k=1}^nm_k\qquad\text{ with }\qquad m_1\geq m_2\geq\cdots\geq m_n.$$
In particular the nonzero $m_k$ form a partition of $n$. Conversely, if $n=a_1+\cdots+a_{\ell}$ is a partition of $n$ then $\ell\leq n$, and sorting the $a_i$ in nonincreasing order and setting $a_k=0$ for all $k>\ell$ we get
$$n=\sum_{k=1}^na_k\qquad\text{ with }\qquad a_1\geq a_2\geq\cdots\geq a_n.$$
So the function of $n$ you are looking for is the partition function. Much is known about it, but the simplest known exact formula is rather scary.
To answer your question; in combinatorics, more often than not a problem has no analytic solution. Rather one could be surprised by the fact that some problems do have analytic solutions.
More specifically for the partition number; counting partitions of a number becomes very messy very fast. Simply listing them all leads to a lot of duplicates such as 
$$4=3+1=2+2=2+1+1=1+3=1+2+1=1+1+2=1+1+1+1,$$
and counting the number of duplicates of each partition does not lead to any nice solution. Alternatively, attempting to count them recursively quickly leads to the expression
$$p(n)=\sum_{k=1}^np_k(n-k),$$
where $p(n)$ denotes the number of partitions of $n$, and $p_k(n)$ denotes the number of partitions of $n$ into parts no greater than $k$. And computing $p_k(n)$, as may be intuitive, is again hard.
