Is 1+2 and 2+1 considered as different ways?

FunforMath wrote: Is 1+2 and 2+1 considered as different ways? These are same ways

Could you post the other problems from the contest?

Generalize slightly, replace $100$ by $n\geqslant 3$. We prove the statement by induction on $n$, postponing base case $n=4$ to the last part. For any nonnegative integer $i\leqslant n-1$. Number of ways to expand $n!$ as sum of elements of $\{1!,2!,...,n!\}$ with exactly $i$ $(n-1)!$'s is greater than $(n-1)!$. To see this, note after removing $i$ $(n-1)!$'s, the remaining sum is $(n-i)(n-1)!$. Thus if we fix a way to expand $(n-i-1)(n-1)!$ then by induction hypothesis, $(n-1)!$ still have at least $(n-1)!$ ways to expand. Since we have divided into $n$ cases, each have more than $(n-1)!$ ways. We are done. To prove the base case, we bash it manually. If there are no $3!$, then we have $12$ ways. If there are one $3!$, then we have $9$ ways. If there are two $3!$'s, then we have $6$ ways. If there are three $3!$'s, then we have $3$ ways. If there are four $3!$'s, then we have $1$ ways. Hence there are total $12+9+6+3+1=31>4!$ ways so we are done. Anyone has Problem 2,3.

The number of ways to express $100!$ as sum of elements of the set $\{1!,2!\}$ is $\frac{100!}{2}$ The number of ways to express $100!$ as sum of elements of the set $\{1!,3!\}$ is $\frac{100!-\frac{100!}{3!}}{5}$ The number of ways to express $100!$ as sum of elements of the set $\{2!,3!\}$ is $\frac{\frac{100!}{2!}-\frac{100!}{3!}}{2}$ The number of ways to express $100!-2$ as sum of elements of the set $\{2!,3!\}$ is about $\frac{\frac{100!}{2!}-\frac{100!}{3!}}{2}$

My solution similar to rmtf1111 's solution. But I think my teacher's idea is very nice: $2 \cdot 1!+2 \cdot 2!+...+n \cdot n!=(n+1)!$

Induct on $n$. For $n=4$, check that $(n-1)!$ can be used at most $4$ times, and for each, a simple counting gives far more than $24$ cases. Suppose the claim is true for $n$. For $(n+1)!$, $n!$ can be used $k$ times at most, where $0\leq k \leq n+1$. For each $k$, the remaining number is $n!(n+1-k)$, and it is straightforward to see that, any writing for $n!$ can trivially be such a writing for $n!(n+1-k)$, e.g. by repeating everything $n+1-k$ many times. Thus, for each fixed $k$, if $n!$ is used $k$ times, we see at least $n!$ ways of partitioning, together with the fact that range of $k$ contains at least $n+1$ distinct integers yields the claim.

Proof by induction. Let denote $S(n)$ as the number of ways expressing the $n!$. For $n=1$, it is obvious. For $n-1$ let assume $S(n-1)\geq(n-1)!$ and $a_{1}1!+a_{2}2!+...+a_{n-2}(n-2)!=(n-1)!$. And we need to prove that $S(n)\geq n!$ for which $a_{1}1!+a_{2}2!+...+a_{n-2}(n-2)!+a_{n-1}(n-1)!=n!$. For each cases below $(n-1)!$ occurs exactly $a_{n-1}$ times. Take $a_{n-1}=0$, then $n[a_{1}1!+a_{2}2!+...+a_{n-2}(n-2)!]+0(n-1)!=n!$ and $(number \, of \, ways) \geq (n-1)!$ Take $a_{n-1}=1$, then $(n-1)[a_{1}1!+a_{2}2!+...+a_{n-2}(n-2)!]+1(n-1)!=n!$ and $(number \, of \, ways) \geq (n-1)!$ $...$ Take $a_{n-1}=n-1$, then $1[a_{1}1!+a_{2}2!+...+a_{n-2}(n-2)!]+(n-1)(n-1)!=n!$ and $(number \, of \, ways) \geq (n-1)!$ After summing up all $(number \, of \, ways)$ we get $S(n)\geq n(n-1)!=n!$ as desired to show.

Please, post all problems from http://matol.kz/olympiads/886 into International Zhautykov Olympiad 2019 section.

Denote $b_n$ to be the total number of ways for $n!$ to be represented as a sum of other factorials. It is easily seen that $b_4 > 24=4!$. Then by recursion we are going to have that: $$b_5 > 5.4!=5!$$this holds since we can represent $5! = 4!+4!+4!+4!+4!$, and we define expanding a factorial to be disintegrating the factorial into the sum of factorials. By this way we have that $b_5 > 5.4!=5!$. Now by recursion we have that $b_{100} > 100!$

very easy problem for IZHO P1

lazizbek42 wrote: very easy problem for IZHO P1 i don't agree with you because all P1 problems were easy especially last years'

We replace $100$ by $n$ and induct on $n$. $n = 4$ is obvious (after casework, that is). Now for any $n+1$, write $(n+1)! = n! + n! + \dots + n!$, where there are $n+1$ factorials on the RHS. Choose some number of factorials (from $1$ to $n+1$), and break all of these down into the same partition of $n!$. This gives at least $(n+1)n! = (n+1)!$ ways, so we're done.