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
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B
132
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C
198
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D
none of these
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Solution
The correct option is B132 Let the numbers selected be x1,x2,x3.
Then we must have : 2x2=x1+x3 ⇒x1+x3= even
Therefore, x1 and x3 both are odd or both are even.
Case I - If x1 and x3 both are even, we can select them in 12C2 ways.
Case II - If x1 and x3 both are odd, we can again select them in 12C2 ways.
Thus, the total number of ways =2×12C2=132