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Answer: $t=3$ and $(x,y,z)$ is any permutation of $(1,2,3)$. Since the equations are symmetrical with respect to variables $x$, $y$ and $z$, we can assume that $x\le y\le z$. Dividing the second equation by the first one we obtain the equality $$t+1=\frac{(t+1)!}{t!}=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right).(1)$$ Assume that $y\ge 3$, then also $z\ge 3$. Now from $x\ge 1$ and (1) we get that $t+1\le 2\cdot \frac{4}{3}\cdot \frac{4}{3}<4$, therefore $t\le 2$, but that contradicts the inequality $y\ge 3$. Thus $y\le 2$. It leaves us with three possibilities. $x=y=1$. Writing (1) in the form $t+1=4+\frac{4}{z}$ we conclude that $z\in \{1,2,4\}$, but none of these values leads to a solution. $x=1$ and $y=2$. From (1) we get that $$t+1=2\cdot \frac{3}{2}\left(1+\frac{1}{z}\right)=3+\frac{3}{z},$$ what means that $z\in \{1,3\}$ and as $z\ge y\ge 2$ then $z=3$ and $t=3$. One can check, that this is a solution, therefore we get 6 solutions where $t=3$ and $(x,y,z)$ is any permutation of $(1,2,3)$. * $x=y=2$. From (1) we get that $t+1=\frac{9}{4}+\frac{9}{4z}$. The expression of the right hand side is larger than 2 and less than 3 if $z\ge 4$, what is impossible. Therefore $z\le 3$ and it remains to check that $(x,y,z)=(2,2,2)$ and $(x,y,z)=(2,2,3)$ are not solutions.