Out of 2n + 1 tickets, which are consecutively numbered, three are drawn at random. Then the probability that the numbers on them are in the arithmetic progression is

Out of 2n + 1 tickets, which are consecutively numbered, three are dra

1.	Out of 2n + 1 tickets, which are consecutively numbered, three are drawn at random. Then the probability that the numbers on them are in the arithmetic progression is
A.	\(\dfrac{n^2}{4n^2 - 1}\)
B.	\(\dfrac{n}{4n^2 - 1}\)
C.	\(\dfrac{3n^2}{4n^2 - 1}\)
D.	\(\dfrac{3n}{4n^2 - 1}\)
Answer» E.

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Out of 2n + 1 tickets, which are consecutively numbered, three are drawn at random. Then the probability that the numbers on them are in the arithmetic progression is

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