Question

Out of 3n consecutive numbers in which three are selected at random, the number of ways such that their sum is divisible by 3 is

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

Answer 1

The correct option is B $\frac{n(3n^2-3n+2)}{2}$

Let the numbers be written as follows:
$3k:3,6,9,12,...3n-------\left(i\right)$
$3k-1:2,5,8,11,...3n-1-----\left(ii\right)$
$3k-2:1,4,7,10,...3n-2-----\left(iii\right)$
Each of the above 3 series has n terms.
Case A: select 3 terms from only series (i) or (ii) or (iii) in: ${}^n{C}_3\times 3=\frac{n(n-1)(n-2)}{2}$
Case B: select 1 term from each of the 3 series in: ${{(}^n{C}_1)}^3=n^3$
Hence, required value = $\frac{n(n-1)(n-2)}{2}+n^3=\frac{n(3n^2-3n+2)}{2}$
Hence, (B) is correct.