Selection of three numbers from $0$ to $11$ with sum $12$ If  $12$  tickets  numbered  $0,  1,  2, \ldots 11$  are  placed  in  a  bag,  and  three  are  drawn  out,  show that  the  chance that  the  sum  of  the  numbers  on  them  is  equal  to  12  is   $\frac{3n}{(6n-1)(6n-2)}=\frac{3}{55}$
My approach:
Let us select three number from $12$: $\binom{12}{3}=220$
Now  I use the concept, I am taking $12$ initially
$$ x_1+ x_2+ x_3=12$$
$x_1, x_2, x_3$  are non-negative.
Number of Cases are $\binom{12+3-1}{3-1}=\binom{14}{2}=91$
Remove three case $(12,0,0),(0,12,0)$ and $(0,0,12)$, as I have taken $12$ which is not required. Number of Cases are $91-3=88$
My answer is $$\frac{88}{220}=\frac{2}{5}$$
Please help me.
 A: You have already counted that the number of draws with a sum of 12, with only condition being all the numbers are between 0 and 12 inclusive, is $\binom{14}{2}$. I'll keep it a bit more general, proving for any number that is a multiple of 6. So instead of 12 I write $6n$, and in your case $n=2$. In this more general case, the formula becomes
$$ \binom{6n+2}{2} = \frac{(6n+2)(6n+1)}{2} = (3n+1)(6n+1) $$
Now remove first the ones that involve repetitions. How many ways are there for the first two to be the same? This is a draw of $(i,i,6n-2i)$, so it can only happen if $0 \leq 6n-2i \leq 6n$ which gives $0 \leq i \leq 3n$. Those are $3n+1$ draws. By symmetry, there are also $3n+1$ draws where the second and third, and the first and third number are the same, respectively. In total, we remove $3(3n+1)$ from the original count, and that also takes care of all three $0,0,6n$ (in any order) cases.
But, we have removed $(2n, 2n, 2n)$ three times (once in each group above), so add 2 to compensate. We now have
$$ (3n+1)(6n+1) - 3(3n+1) + 2 = 18n^2$$
ways to draw three different numbers. Since we don't care about the order though, we have counted each unique draw 6 times (3P3) so divide by 6 to get $3n^2$.
Combine that with the total (which you already had for the 12 case), and you get probability
$$ \frac{3n^2}{\binom{6n}{3}} = \frac{3n}{(6n-1)(6n-2)} $$
