NIMCET 2018 Question Paper with solutions

Total number of ways of choosing 3 numbers from the given set of numbers 1, 2,...99 = $\stackrel{99}{}{\mathit{C}}_3$
Let us divide the given numbers into 3 groups ${\mathit{G}}_1,{\mathit{G}}_{2.}{\mathit{G}}_3$ as follows
${\mathit{G}}_1$ : 3 , 6 , 9 , ... , 99 [33 elements]
${\mathit{G}}_2$ : 1 , 4,7...97 [33 elements]
${\mathit{G}}_3$ : 2,5,8... 98 [33 elements]
Given that ${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3$ - 3abc is to be divisible by 3.
If ${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3$ - 3abc is divisible by 3, then which is possiblein the following cases.
(i) All the numbers belong to the first group.(ii) All the numbers belong to the second group
(iii) All the numbers belong to the third group
(iv) One number belong to the first group, one belong to second group and one belong to the third group.
So, favourable number of cases are 3. $\stackrel{33}{}{\mathit{C}}_3+{\left(\stackrel{33}{}{\mathit{C}}_1\right)}^3$
∴ Required probability = $\frac{3.\stackrel{33}{}{\mathit{C}}_3+{\left(\stackrel{33}{}{\mathit{C}}_1\right)}^3}{\stackrel{99}{}{\mathit{C}}_3}$