Out of (2n+1) tickets numbered consecutively, three are drawn at random. the chance that the numbers on them are in A.P is
(a) n24n2-1
(b) 2n4n2-1
(c) 3n4n2-1
(d) 4n4n2-1
865The number of selections without the constraint is 2n+1C3. Now, for common difference=1, there are (2n-1) possible selections, for c.d.=2, there are (2n-3) selections and so on. Thus, the total number of allowed selections is (2n-1)+(2n-3)+...+5+3+1=n2. Therefore the probability is 3n4n2-1.
Suparno Ghoshal father u r awesome!!!!!!!!!!!!!
591Another way is that for 3 numbers in AP
c-a=2b ...so the difference btwn 1st and 3rd is even
so if we choose the 1st number to be even then the 3rd shud also b even and the 2nd will be automatically selected..
now total even nos.=n ..so no. Of ways of choosing 2 nos.=nC2
for odd= n+1C2
and total no. Of possibilities= 2n+1C3
so ,P= nC2+(n+1)C2/(2n+1)C3..simplifying u get (c)
Soumyadeep Basu Yeah, absolutely.
37Yeah that is one option. But isn't there some other way of solving it ?
