Prove there are at least $100!$ ways to write $100!$ as a summation of numbers in $\{1!,2!,...,99!\}$. (each number can be used multiple times.) Prove there are at least $100!$ ways to write $100!$ as a summation of numbers in $\{1!,2!,...,99!\}$. (each number can be used multiple times.)
I have no idea how to solve this question. First, I thought of using mathematical induction and seeing the reason for $2!$ and then trying to find out a pattern for the rest, but it did not work.
For $3!$ in set $\{1!,2!\}$ I can do the addition like:
$1+1++1+1+1$
$1+1+1+1+2$
$1+1+2+2$
$2+2+2$
and so it failed for this one too and I could not use induction.
If someone could help me with this I would really appreciate it.
 A: If $n$ is even, then there are $1+\frac{n}{2}$ ways to write it as a sum of $1$ and $2$.

If $n$ is divisible by $3!=6$, then how many ways are there to write it as a sum of $1$, $2$, and $6$?
In any sum, there are $k$ $6$'s, where $0 \leq 6k \leq n$, and the number $n-6k$ is necessarily even, so there are $$1 + \frac{n-6k}{2}$$ ways to write $n-6k$ using only $1$ and $2$.  Therefore, there will be
$$\sum_{k=0}^{n/6}\left(1 + \frac{n-6k}{2}\right) = \left(1 + \frac{n}{6}\right)\left(1+\frac{n}{4}\right)$$
ways to write $n$ as a sum of $1$, $2$, and $6$.

Now, it's useful to note that $$n \geq 18 \implies \left(1 + \frac{n}{6}\right)\left(1+\frac{n}{4}\right) > n$$
so for any $n \geq 18$, there will be more than $n$ ways to write $n$ as a sum of $1$, $2$, and $6$.
In particular, $100! \geq 18$, so there are many more than $100!$ ways to write $100!$ as a sum of $1$, $2$, and $6$, so many more than $100!$ ways to write it as a sum using $1!, 2!, 3!, \ldots, 99!$.
A: Assumed that question intends at least $100!$ ways.  Also assumed that order of terms deemed irrelevant.  That is for $N = 4$ it is presumed that if there were $6$ terms, that consisted of $3$ terms of $3!$ and $3$ terms of $2!$ that the order that the $6$ terms appeared in the summation would be deemed irrelevant.
Use induction:
Base Case : $N = 4$.
The number of $3!$ terms can be : 
$0$ : $13$ possibilities for the number of $2!$ terms. 
$1$ : $10$ possibilities for the number of $2!$ terms. 
$2$ : $7$ possibilities for the number of $2!$ terms. 
$3$ : $4$ possibilities for the number of $2!$ terms. 
$4$ : $1$ possibilities for the number of $2!$ terms.
Suppose conjecture true for $N \geq 4$.
In summing terms to add to $(N+1)!$, the number of $N!$ terms can be:
$0:$ 
$1:$ 
$2:$ 
$\cdots$ 
$(N):$ 
$(N+1):$ 
Of the $(N+2)$ possible number of $N!$ terms that will be used, in all but the very last possibility, the corresponding deficit will always be $\geq (N!)$.
Using the induction hypothesis, anytime that the deficit is $\geq (N!)$, that deficit can be completed in at least $(N!)$ ways, through the use of the terms $\{1!, 2!, \cdots, (N-1)!\}$.  Therefore, the number of ways of completing the summation to $(N+1)!$ must be strictly greater than $(N+1) \times N!$.
