Odds of specific scenario after $4$ dice rolls I have started studying for a statistics competition but seem to be a bit rusty and unsure about the answer (which I can't check). 

"A fair die is rolled $4$ times. What's the probability of each of the numbers $1, 2$ and $3$ appearing at least once each?"

I tried thinking that I can get the numbers $1, 2$ and $3$ then a fourth roll providing $6$ different numbers. That would leave me $6$ possibilities of getting $1, 2$ and $3$ on the $3$ first rolls, in that order.

Since I don't need the $1,2$ and $3$ necessarily on the $3$ first rolls, I thought they can actually be placed in $C_{4,3} =4$ "positions" of rolls.

And also, they can come in $3!=6$ possible orders. 

Adding all that up equals to ${\rm{6}}{\rm{.4}}{\rm{.6 = 144}}$ possibilities of getting $1, 2$ and $3$ at least once, from a total of ${6^4} = 1296$ different outcomes.

$$P = {{144} \over {1296}} \approx 0.11$$

It is a multiple choice and the answer is one from: $${5 \over {36}},{1 \over {12}},{5 \over {12}},{1 \over {24}}$$

And none of the above is equivalent to 0.11, unfortunately. Can anyone help figuring out which step I got wrong?
Brute-forcing:
$$\begin{array}{c|c|c|c}
123(6)&amp;12(5)3&amp;1(4)23&amp;(3)123\\
\hline
132(6)&amp;13(5)2&amp;1(4)32&amp;(3)132\\
\hline
213(6)&amp;21(5)3&amp;2(4)13&amp;(3)213\\
\hline
231(6)&amp;23(5)1&amp;2(4)31&amp;(3)231\\
\hline
312(6)&amp;31(5)2&amp;3(4)12&amp;(3)312\\
\hline
321(6)&amp;32(5)1&amp;3(4)21&amp;(3)321\\
\hline
3!\cdot 6&amp;3!\cdot 5&amp;3!\cdot 4&amp;3!\cdot 3&amp;108\end{array}$$
Hence, the required probability is:
$$\frac{108}{6^4}=\frac1{12}.$$
Let's just count the number of ways it could happen.

If they each appear once each, so the last number is a 4,5,6:

* pick if the last number is 4,5,6 (3)

* order everything (4!)


If one of them appears twice:

* pick which appears twice (3)

* order everything, but unorder the numbers that appear twice (4!/2!)


Add those products and then divide by 6^4.