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 $b \neq c$. 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 $\max\{b, c\} \geq a+1$. 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.