# 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.

• So it is drawing without replacement? Then no two of the three can be the same. Commented Oct 2, 2017 at 11:24
• You have included the drawing of $(4, 4, 4)$ and $(3, 3, 6)$, for instance, in your count of draws that sum to $12$. That is probably not your intention, seeing as the total number of ways is described as $\binom{12}3$, which implies that the same number cannot be drawn twice. Also, you have counted $(3, 4, 5)$ as a different drawing from $(5, 4, 3)$ in your second counting, but not in your first. Commented Oct 2, 2017 at 11:25
• it all depends on with replacement or without, with fixed points or without ... etc.
– user451844
Commented Oct 2, 2017 at 11:30
• Thanks i did thought of it,i can manually calculate the number of cases which is 12, {11,1,0}, {10,2,0},{9,3,0},{8,4,0}, {7, 5,0}, {9,2,1},{8,3,1},{7,4,1},{6,5,1},{7,3,2},{6,4,2}&(4,3,5}=12 cases Commented Oct 2, 2017 at 11:32
• What is the meaning of $n$ in the given formula ? Commented Oct 2, 2017 at 12:08

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 $$\frac{3n^2}{\binom{6n}{3}} = \frac{3n}{(6n-1)(6n-2)}$$