Mathematics - Permutation & Combination Question with Solution | TestHub

$\mathrm{a}+\mathrm{b}+\mathrm{c}=21\Rightarrow \mathrm{b}+\mathrm{c}>\mathrm{a}\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}>2\mathrm{a}\Rightarrow 2\mathrm{a}<21\Rightarrow \mathrm{a}\le 10$.

So $1\le \mathrm{a},\mathrm{b},\mathrm{c}\le 10$. The cases when $\mathrm{a}>\mathrm{b}>\mathrm{c}$ are $(10,9,2),(10,8,3),(10,7,4),(10,6,5),(9,8,4),(9)$, 7, 5) and (8, 7, 6). So, number of cases when $\mathrm{a},\mathrm{b},\mathrm{c}$ are all distinct is $7\times 3!=42$. The cases when $\mathrm{a}=\mathrm{b}>\mathrm{c}$ or $\mathrm{a}>\mathrm{b}=\mathrm{c}$ are $(10,10,1),(9,9,3),(8,8,5)$ and $(9,6,6)$. So number of cases when two same and 1 different is $4\times 3!/2!=12$. The cases when $\mathrm{a}=\mathrm{b}=\mathrm{c}$ is (7,7,7). The total number of ordered triplets $=42+12+1=55$.