Question

The total number of ways in which three
distinct numbers in $A.P$ can be selected from the set $\{1,2,3,\cdots ,24\}$ is equal to :

• A. $66$
• B. $132$
• C. $198$
• D. none of these

Answer 1

The correct option is B $132$

Let the numbers selected be $x_1,x_2,x_3.$
Then we must have :
$2x_2=x_1+x_3$
$\Rightarrow x_1+x_3=\text{even}$
Therefore $x_1,x_3$ both are odd or both are even.
Case I - If $x_1\text{and}{x}_3$ both are even, we can select them in ${}^{12}{C}_2$ ways.
Case II - if $x_1\text{and}{x}_3$ both are odd, we can again select them in ${}^{12}{C}_2$ ways.
Thus, the total number of ways: $=2\times {}^{12}{C}_2=132.$