Vtyjkyuk

Find the number of ways of selecting n letters from 3n letters which contains 'n' a s , 'n' b s and the rest n letters are distinct from each other.

From the language of the problem I can easily understood that we need to select n letters.

I will categorize it into three category using
$C(n,r) = fracn!(n-r)!r!$

I will use $C(n, r_1)*C(n, r_2)*C(n, r_3)$,

$ r_1$ =a group, $ r_2$ =b group & $ r_3$ =unlike group

Where $ r_1+ r_2 + r_3 =n$ I presume that i am on the right approach but I am not able to proceed hence forth.

You've already provided the answer in a comment: $(n+2)2^n-1$.

Here's an algebraic proof: There are $k+1$ ways to select $k$ a's and b's, and then there are $binom nk$ ways to choose the distinct letters you don't select. Thus the desired count is

begineqnarray*
sum_k=0^n(k+1)binom nk
&=&
left[fracpartialpartial qsum_k=0^nbinom nkq^k+1right]_q=1
\
&=&
left[fracpartialpartial qleft(q(1+q)^nright)right]_q=1
\
&=&
2^n+n2^n-1;.
endeqnarray*

Here's a combinatorial proof: The selections without b's correspond to the $2^n$ subsets of the $n$ distinct letters. (Add the right number of a's to such a subset, and you get a unique selection without b's.) Now consider the selections with at least one b. Each contains a proper subset $S$ of the $n$ distinct letters, and the number of elements not contained in $S$ is the number of choices for the non-zero number of b's to add. Thus, the choice of the non-zero number of b's corresponds to distinguishing one of the distinct letters not contained in $S$. Thus, the selections with at least one b are in bijection with the ways to distinguish one of the distinct letters ($n$ choices) and to choose a subset of the remaining distinct letters ($2^n-1$ choices). Hence the total number of selections is $2^n+n2^n-1$.