1708\hspace{-16}$Let $\bf{1-}$ apperar $\bf{x}$ no. of times.\\\\\\ and $\bf{2-}$ apperar $\bf{y}$ no. of times.\\\\\\ and $\bf{3-}$ apperar $\bf{z}$ no. of times.\\\\\\ Now Given in Question::\\\\\\ $\bf{x+2y+3z=10}$ and $\bf{x+y+z=7}$\\\\\\ after solving we Get $\bf{y+2z=3}$\\\\\\ which is possible when $\bf{y=1\;\;,z=1}$ or $\bf{y=3\;,z=0}$\\\\\\ So for first case $\bf{x=5,y=1,z=1}$ means $\bf{\left(1,1,1,1,1,2,3\right)}$\\\\\\ So for second case $\bf{x=4,y=3,z=0}$ means $\bf{\left(1,1,1,1,2,2,2\right)}$\\\\\\ So Total no. is $\bf{=\frac{7!}{5!}+\frac{7!}{4! \times 3!}=42+35=77}$
