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

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Solution

$Let the numbers selected be x_{1}, x_{2}, x_{3}$
$We must have 2 x_{2} = x_{1} + x_{3}$
$\Rightarrow x_{1} + x_{3} = even$
$Therefore both x_{1}, x_{3} are odd (or) both are even.$
$If x_{1} {and x}_{3} {both are even, we can select them in}^{12} C_{2} ways.$
$Similarly, if both x_{1} and x_{3} are odd,$
${we can again select them in}^{12} C_{2} ways.$
$Thus, the total number of ways is 2 \times^{12} C_{2} = 132$

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Combinations

Remember concepts with our Masterclasses.

The total number of ways in which three distinct numbers in Arithmetic Progression can be selected from the set 1,2,3,.......,24 is equal to

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