27. Solve the equation $(a+b) c!=(a!+b!) c$ in natural numbers
27. Answer: $\mathrm{c}=1, \mathrm{a}=1, \mathrm{~b}=2 ; \mathrm{c}=1, \mathrm{a}=2, \mathrm{~b}=1 ; \mathrm{c}=1, \mathrm{a}=2, \mathrm{~b}=2 ; \mathrm{c}=\mathrm{a}=\mathrm{b}-$
any.

Let's assume that $a \geqslant b$. Then there are three cases:

Case 1: $c>a$.

Then $(a+b) c!=(a+b) a! \cdot x \cdot c > 2 \cdot a! \cdot x \cdot c \geqslant (a!+b!) c$ (except for the case $a=b=1, c=2$).

Case 2: $c<a$.

Then $(a+b) c! < a! \cdot c \cdot c! = a \cdot x \cdot c! \cdot c \geq 2 \cdot a \cdot c! \geq (a+b) c!$ (except for the case $c=1$, for which $a!+b!=a+b$, hence $a, b=1$ or 2, but $c<a$, so $c=1=b, a=2$ or $c=1, a=b=2$).

Case 3: $c=a$

Then $(a+b) a! = (a!+b!) a$, expanding the brackets, we get $a \cdot a! + b \cdot a! = a! \cdot a + b! \cdot a$, from which $b \cdot a! = b! \cdot a$, dividing by $a b$ we get: $(a-1)! = (b-1)!$, hence $a=b=c$.