Question

Out of $3n$ consecutive numbers, the number of ways in which $3$ numbers can be selected such that their sums is divisible by $3$ is

• A. $\frac{3n^2-3n+2}{2}.n$
• B. $\frac{3n^2-3n+2}{3}.n$
• C. $\frac{2n^2-3n+3}{3}.n$
• D. $\frac{2n^2-3n+3}{2}.n$

Answer 1

Answer: (A)

$\frac{3n^2-3n+2}{2}.n$

Out of $3n$ consecutive numbers there will be
$n\to 3k$
$n\to 3k+1$
$n\to 3k+2$
Given that sum should be divisible by 3
so we can select a number from each set.
or 3 from same set.
so total no of possible ways are $3\left[\frac{n(n-1)(n-2)}{3!}\right]+n^3$
$\Rightarrow \left[\frac{3n^2-3n+2}{2}\right]n$