Suppose 12 i.i.d. integer samples are taken from 1 to 10 with uniform probability. What is the joint distribution of the 3 last numbers ($x_{10},x_{11},x_{12}$) and the 3 largest numbers ($\max_1(x_1,...,x_{12}),\max_2(x_1,...,x_{12}),\max_3(x_1,...,x_{12})$) in the sample?
$$ p(x_{10},x_{11},x_{12},\max_1(x_1,...,x_{12}),\max_2(x_1,...,x_{12}),\max_3(x_1,...,x_{12}))$$
I think the 3 largest numbers can be modeled using the CDF of uniform(1,10) but how do we account for the last 3 numbers? I am thinking it could involve taking the conditional on the 3 largest numbers to describe the distribution of the last three numbers.
$$ p(x_{10},x_{11},x_{12}|\max_1(x_1,...,x_{12}),\max_2(x_1,...,x_{12}),\max_3(x_1,...,x_{12}))\cdot p(\max_1(x_1,...,x_{12}),\max_2(x_1,...,x_{12}),\max_3(x_1,...,x_{12}))$$
I'll do the joint distribution of the last number and the largest number:
There are but two valid cases: $(1)$ the last number is the largest or $(2)$ it is less than the largest.
$$\begin{align}\mathsf P(X_{12}=z, X_{(12)}=j) ~=~& \begin{cases} \mathsf P(X_{12}=z, \max\limits_{0\leq i\leq 11}(X_i)\leq z) & : 1\leq z=j\leq 10 \\[1ex] \mathsf P(X_{12}=z, \max\limits_{1\leq i\leq 11}(X_i)=j) & : 0\leq z<j\leq 10 \\[1ex] 0 & : \textsf{otherwise} \end{cases} \\[2ex] ~=~& \begin{cases} {j^{11}}/{10^{12}} & : 1\leq z=j\leq 10 \\[1ex] (j^{11}-(j-1)^{11})/10^{12} & : 0\leq z<j\leq 10 \\[1ex] 0 & : \textsf{otherwise} \end{cases}\end{align}$$
Use similar logic to evaluate: $\mathsf P(X_{10}{=}x, X_{11}{=}y, X_{12}{=}z, X_{(10)}{=}k, X_{(11)}{=}h, X_{(12)}{=}j)$
There will be many more cases. (Tip: look for symmetries.)