Out of 2n tickets numbered `1, 2, ......, 2n`, three are chosen at random. The probability that the numbers on them are in A.P. is

Out of 21 tickets numbered 10, 11, 12,…, 30, three tickets are drawn at random. Find the probability that the numbers on these tickets are in A.P.

From 4m+1 tickets numbered as 1,2 ,….4m+1…. Three tickets are chosen at random. Find out the probability that the numbers are in A.P. with even common difference.

Out of 21 ticket marked 1,2….21 , three are drawn at random without replacement . The probability that these numbers are in A.P is :

Out of (2n+1) tickets consecutively numbered , three are drawn at random . The chance that the numbers on them are in A.P is

Out of 15 tokens consecutively numbered from 1 to 15, 3 tokens are drawn at random. The probability that the numbers on them in AP is

Out of (2n +1) tickets cosecutively numbered, three are drawn of random, find the probability that the three number on them are in A.P.

Statement 1: Out of 5 tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is 2//15 . Statement 2: Out of 2n+1 tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in A.P. is 3n//(4n^2-1) .

Statement 1: Our of 5 tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is 2//15 . Statement 2: Out of 2n+1 tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in A.P. is 3n//(4n^2-1) .

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.