Q13/121) Out of ( 2n + 1 ) tickets, consecutively numbered, 3 are drawn at random. Find the chance that numbers on them are in A.P.

The total number of cases (2n+1C3)

The lowest number can be 1 to 2n-1
if the lowest number is odd then there are even number of numbers and 2^nd number shall be in 1^st half.

So if lowest number id 2k +1 there are (n-k) choices for 2^nd and 3^rd number pairs

if the lowest number is even  then there are odd number of numbers and 2^nd number shall be in 1^st half and middle number not counting.

So if lowest number id 2k there are (n-k) choices for 2^nd and 3^rd number pairs

If lowest number is 1 total number of cases n( lower of the rest 2 from 2 to n+ 1
If lowest number is 2 total number of cases n-1
If lowest number is 3 total number of cases n-1
If lowest number is (2n-2) total number of cases 1( that is 2n-1 is the lower of the 2)
If lowest number is (2n-1) total number of cases 1( that is 2n is the lower of the 2)
So total number of cases = n(n-1)/2 *2 + n = n^2
So probability or chances= n^2/(2n+1C3)