IRL_ABooklet

There are trivial solutions when $a=b=c$, and one non-trivial solution when $a=2,b=3,c=1$ as: $$2!=2=\frac{1}{5}\times 6+\frac{4}{5}\times 1=\frac{1}{5}\times 3!+\frac{4}{5}\times 1!$$ We will prove that there are no other solutions. If $b=c$ the equation becomes $a!=b!$ and we must have $a=b$, which gives the trivial solution.

Suppose that $b\ne c$. Then, as $a!$ is a weighted average of $b!$ and $c!$, then $a!$ lies strictly between $b!$ and $c!$. As the factorial is a strictly increasing function, this implies $a$ lies strictly between $b$ and $c$, and in particular $\max \{b,c\}\ge a+1$. This gives $$a!=\frac{b!+4c!}{5}>\frac{b!+c!}{5}>\frac{(a+1)!}{5}=\frac{a+1}{5}a!$$ Dividing by $a!$ and multiplying by 5 gives $5>a+1$ whence $a\in \{1,2,3\}$. The case $a=1$ is excluded, since $a$ strictly exceeds the lesser of the positive integers $b$ and $c$.

The case $a=2$ leads to $b!+4c!=10$, so $c!<\frac{10}{4}$ and $c\le 2$. The sub-case $c=1$ gives the non-trivial solution above, while $c=2$ is not possible as $a$ is strictly between $b$ and $c$.

When $a=3$, we have to solve $b!+4c!=30$. As $c!<\frac{30}{4}$ we must have $c\le 3$. We cannot have $c=3$ because $a$ is strictly between $b$ and $c$. When $c=2$ we should have $b!+8=30$, and when $c=1$ we should have $b!+4=30$, neither of which has a solution.