Question

The total number of ways of selecting two number from the set $\{1,2,3,4,....,3n\}$, so that their sum is divisible by $3$, is equal to

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

Answer 1

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

Given number can be rearranged as
$P=\{1,4,7,....,3n-2\},n\left(P\right)=n$
$Q=\{2,5,8,....,3n-1\},n\left(Q\right)=n$
$R=\{3,6,9,....,3n\},n\left(R\right)=n$

For sum of $2$ numbers to be taken from above sets such that their sum is divisible by $3$, we must take two numbers from $R$ or one number each from $P$ and $Q$.
$\therefore$the total number of required ways is
${}^n{C}_2+({}^n{C}_1\times {}^n{C}_1)=\frac{n(n-1)}{2}+n^2=\frac{3n^2-n}{2}$