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

Answer: (B)

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

Any number divisible by $3$ can be written as $3\lambda -2,3\lambda -1,3\lambda$

Arranging $\{1,2,3,4,...,3n\}$ as

$A=\{1,4,7,...,3n-2\}$ numbers of the form $3\lambda -2$

$B=\{2,5,8,...,3n-1\}$ numbers of the form $3\lambda -1$

$C=\{3,6,9,...,3n\}$ numbers of the form $3\lambda$

$\therefore$ there are three sets of numbers .

Let us select $2$ numbers from set $C$ and $1$ number from set $B$ and $1$ number from set $A$

${=}^n{C}_2{+}^n{C}_1{\times }^n{C}_1$

$=\frac{n!}{(n-2)!2!}+\frac{n!}{(n-1)!}\times \frac{n!}{(n-1)!}$

$=\frac{n(n-1)(n-2)!}{(n-2)!2!}+\frac{n(n-1)!}{(n-1)!}\times \frac{n(n-1)!}{(n-1)!}$

$=\frac{n(n-1)}{2}+n^2$

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