From $4m+1$ tickets numbered as $1, 2, \ldots 4m+1$. Three tickets are chosen at random. The probability that the numbers are in A.P. with even common difference is
$\frac{2(2m-1)}{3(16m^2-1)}$
$\frac{3(2m+1)}{2(16m^2+1)}$
$\frac{3(2m-1)}{2(4m^2-1)}$
None of these
Step-by-Step Solution
Key Concept: Count the number of arithmetic progressions with common difference $d$ by systematically varying the starting term, then divide by total combinations.
Let $d$ be the common difference. For $d=2$, we can form triplets like $(1,3,5), (2,4,6), \ldots, (4m-3, 4m-1, 4m+1)$, giving $4m-3$ triplets. For $d=4$, we get $4m-7$ triplets. Continuing this pattern, the total number of favorable triplets is $m(2m-1)$. The total ways to select 3 numbers from $4m+1$ numbers is $^{4m+1}C_3$. The required probability is $\frac{m(2m-1)}{^{4m+1}C_3} = \frac{3(2m-1)}{2(16m^2-1)}$.
Correct Answer: 4