Upper bound on the number of ways to completely fill boxes. I have $x$ indistinguishable items and $n$ boxes with sizes $y_1,\dots,y_n$. I am interested in the number of ways of allocating those $x$ items into the boxes such that all boxes must be either fully filled (i.e., the number of items in each box equals its size) or empty. Also, I have the condition $x=o(n)$ (where $x\leq n/2$ is included in the condition), which indicate that boxes of smaller sizes might possibly lead to more number of ways.
For example, if I have 3 items and 100 boxes all with size 2, then I have 0 ways to achieve my goal. However, if one of the boxes have size 3, then I would have one way to achieve my goal. If I have 3 items and 100 boxes all with size 1, then I have ${100\choose 3}$ ways to achieve my goal.
I suspect that the upper bound for the number of ways of allocating those $x$ items into $n$ boxes such that all boxes must be fully filled is ${n\choose x}$, which is the case where all boxes have size 1. I'm having trouble arguing this rigorously. Is there a way to show this rigorously or is the claim wrong?
 A: A generating function argument is that the number of ways of doing this is the coefficient of $z^x$ in the expansion of $\left(1+z^{y_1}\right)\left(1+z^{y_2}\right)\left(1+z^{y_3}\right)\cdots \left(1+z^{y_n}\right)$
That is certainly less than or equal to the coefficient of $z^x$ in the expansion of $\left(1+z\right)^{\sum y_i}$, which is ${\sum y_i \choose x}$, but you have in effect asked if it is always less than or equal to the coefficient of $z^x$ in the expansion of $\left(1+z\right)^{n}$, which is ${n \choose  x}$.
The answer to that could clearly be no in some cases even when $x<n$.  For example if $n=10$ and $x=8$ and each $y_i=2$ then there are ${10 \choose 4}= 210$ possibilities in the box size $2$ case and ${10 \choose 8}= 45$ possibilities in the box size $1$ case.
So it may come down to precisely what you mean by $x=o(n)$.
A: Just to flesh out the details of the proof alluded to in the comments:
Let $\mathcal A$ be the collection of $S$ for which $\sum_{i\in S}y_i=x$. Then $\mathcal A$ is an antichain in the poset of subsets of $\{1,\dots,n\}$. Let $a_k$ be the number of subsets of size $k$ in $\mathcal A$.
The LYM inequality implies
$$
\sum_{k=0}^n \frac{a_k}{\binom{n}k}\le 1.\tag{*}
$$
Therefore,
$$
|\mathcal A|
= \sum_{k=0}^n a_k
\stackrel{1}=\sum_{k=0}^{x}a_k
= \binom{n}{x}\sum_{k=0}^{x}\frac{a_k}{\binom{n}x}
\stackrel{2}\le \binom{n}{x}\sum_{k=0}^{x}\frac{a_k}{\binom{n}k}
\stackrel{*}\le \binom{n}x\cdot 1.
$$

*

*We must have $a_k=0$ for $k>x$, since if there were more then $x$ full boxes, the total number of items would be more than $x$.


*Here, we use the fact that $x<n/2$, so that $\binom{n}k\le \binom{n}x$ for all $k\le x$.
