How many combinations possible with order rules? I have these base values "$a_1,a_2,a_3,a_4,b_1,b_2,b_3,b_4,c_1,c_2,c_3,c_4$" and I have to find how many different ways I can arrange them. Rules are these, for example $a_2$, can't come before $a_1$, all a's have to be in order. same goes for b's and c's. But for example $b_1$ or $c_1$ can come before $a_2$ etc.
Here is incorrect line: $a_1,b_1$,$a_3$, $a_2$,$b_2,b_3,b_4,c_1,c_2,c_3,c_4,a_4$ // $a_3$ comes before $a_2$, so a's are not in correct order.
Here is correct line: $b_1,a_1,a_2,c_1,a_3,b_2,a_4,c_2,b_3,b_4,c_3,c_4$ // everything is fine.
I'd like to know is there any formulas that can solve this?
 A: Take $4$ red sticks, $4$ blue sticks and $4$ yellow sticks and arrange them in a row of $12$ sticks. 
The positions for the red sticks can be chosen in $\binom {12}{4}$ ways. 
The blue sticks have to be fitted into the eight positions which remain, which can be done in $\binom 84$ ways, and the yellow sticks fit in the four remaining places. This is $\binom 44$ ways if you like to be neat. 
The first red stick is the position for $a_1$, the second for $a_2$ etc. The blue sticks locate the $b_i$ and the yellow sticks the $c_i$. It is easy to see there is a bijection between the two kinds of arrangement - colours and letters - each determines the other.
So the answer is $$\binom {12}{4}\binom 84\binom 44=\binom {12}{4}\binom 84$$
A: There are $12$ symboles, so $12!$ possibilities to order them without taking into account rules. Let $A$ be the set of all these sequences.
Let $S_4$ be the group of permutations of 4 elements. $(S_4)^3$ acts on $A$ in the following way:
For example, let $\sigma=(\sigma_a,\sigma_b,\sigma_c)\in(S_4)^3$, then $$\sigma(b_1,a_1,a_2,c_1,a_3,b_2,a_4,c_2,b_3,b_4,c_3,c_4)
=b_{\sigma_b (1)},a_{\sigma_a (1)},a_{\sigma_a (2)},c_{\sigma_c (1)},a_{\sigma_a (3)},b_{\sigma_b (2)},a_{\sigma_a (4)},c_{\sigma_c (1)},b_{\sigma_b (3)},b_{\sigma_b (4)},c_{\sigma_c (3)},c_{\sigma_c (4)}.$$
The correct sequences are representant of each orbit of this group action.
So the number of correct sequences is the number of orbits:
 $$\mathrm{Card}(A)/\mathrm{Card}((S_4)^3)=12!/((4!)^3)=34\ 650.$$
