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

Explanation

This offers an elegant way of solving the problem. Since numbers `a_1, a_2, a_3` are in AP.
`implies 2a_2=a_1+a_3`
Now LHS is even, so is RHS. For `a_1+a_3` to be even, `a_1 " and " a_3` must have the same parity - that is, they are either both odd or both even.
Out of (2n+1) tickets consecutively numbered, either (n+1) of them will be odd and n of them even or n of then will be odd and (n+1) of them even.
`therefore n(E)=""^(n+1)C_2+""^nC_2=((n+1)n)/2+(n(n-1))/2=n/(2).2n=n^2`
Required probability `=(n(E))/(n(S))=(n^2)/((n(4n^2-1))/3)=(3n)/(4n^2-1)`