Mathematics - Permutation & Combination Question with Solution | TestHub

$a+b+c=21\Rightarrow b+c>a\Rightarrow a+b+c>2a\Rightarrow 2a<21\Rightarrow a\le 10$. So $1\le a,b,c\le 10$. The cases when $a>b>c$ 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 a, b, c are all distinct is $7\times 3!=42$. The cases when $a=b>c$ or $a>b=c$ 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 $=42+12+1=55$.