Question

The total number of ways in which three distinct numbers in A.P. can be selected from the set $\{1,2,3,...,24\}$ 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 $x_1,x_2,x_3$. We must have
$2x_2=x_1+x_3$
$\Rightarrow x_1+x_3=even$.
Therefore, $x_1,x_3$ both are odd or both are even.
If $x_1$ and $x_3$ both are even, we can select them in ${}^{12}{C}_2$ ways.
Similarly,
if $x_1$ and $x_3$ both are odd, we can again select them in
${}^{12}{C}_2$ ways.
Thus, the total number of ways is $2{\times }^{12}{C}_2=132$.