Very easy..... WLOG x≥y . Hence y≤5.
Easy computation gives no solution for y=5.
If y≤x≤5. we get tiplets:
\boxed{\left(x,y,z \right)=\left(3,4,1 \right)}
\hspace{-16}$Find all Positive Integer Triplet $\bf{(x,y,z)}$ that satisfy\\\\ $\bf{x!+y!=15\times 2^{z!}}$
Very easy..... WLOG x≥y . Hence y≤5.
Easy computation gives no solution for y=5.
If y≤x≤5. we get tiplets:
\boxed{\left(x,y,z \right)=\left(3,4,1 \right)}