12 students participate in 3 competitions twice in total. Count possible results. There are 12 students, who participate in 3 competitions. 
Every student participate twice (in total, once in 2 competitions or twice in 1). 
Count how many possible results of competitions exist -  how many possible 3 ranking lists permutations exist. 
We assume that there are no draws, so result of every competition is a ranking list (can be empty or contain repetitions). 
I know how to this when every result of student is distinct - then it would be just $\binom{24 + 3 - 1}{3-1} * 24!$.
 But in this case we must somehow subtract possibilities when we swap the same person. 
Can you give me some clues?
Example: 

For 2 students(John, Anne) possible ranking lists are:

1. |John|Anne|John|  - John participates in 1st and 3rd competition
   |Anne|    |    |  - Anne participates in 1st and 2nd competition

2. |Anne|Anne|John|  - John participates in 1st and 3rd competition
   |John|    |    |  - Anne participates in 1st and 2nd competition

3. |John|Anne|Anne|  - John participates in 1st competition twice
   |John|    |    |  - Anne participates in 2nd and 3rd competition

4. |John|    |Anne|  - John participates in 1st competition twice
   |John|    |Anne|  - Anne participates in 3rd competition twice
...


 A: Actually after a few days I found the answer (a closed formula). 
It's a modification of the solution with distinct results. 
Let's take $12$ students, each one of them with $2$ chances to participate, and put their names on a ball once for each of their chances.
So we have $24$ balls with names of $12$ students (each name twice). 
Now we put balls in row in any order.
So now let's take $2$ barriers to divide balls onto $3$ sets:
**...*| **...*|**...* .
We want to know how many possibilities of such distribution. 
When we put barriers next to the balls :
**..*||  we see that we have $24+2=26$ positions and we have to choose $2$ from $14$ positions for barriers -> so we have $\binom{26}{2}$ possibilities.  
So now we know that there are $\binom{26}{2}$ possibilities of dividing students onto $3$ competitions.
But we want to consider order in this distributions, because we distinguish order in the ranking list.
So we multiply $\binom{26}{2}$ by $24!$ because there are $24!$ possibilities of balls distribution.
But some of balls are indistinguishable, because there is a copy of every ball for a second chance of student.
So when let's choose position for first ball(f1) and it's copy(c1).
f1 _ c1 _ _... - we chose 1st position for f1 and 3rd for c1
c1 _ f1 _ _... - we chose 3rd position for f1 and 1st for c1
We count this as $2$ possibilities, but from our point of view
f1 _ f1 _ _... 
f1 _ f1 _ _... 
ball and it's copy are the same ball.
To get rid of the redundant possibilities we will divide our result by $2$ for every distinct ball, because earlier for every two positions we swapped indistinguishable balls once and got another possibility.
So the final answer is: $$\frac{\binom{26}{2} * 24!}{2^{12}} = 49\ 229\ 914\ 688\ 306\ 352\ 000\ 000 $$
