Permutation of coefficient with coditions I have 6 coefficients, (V1,V2,H1,H2,D1,D2). Their permutation is 6! = 720.
But I have a rule: V2 cannot lead V1, H2 cannot lead H1 and D2 cannot lead D1.
For example:
V2V1H1H2D1D2 is prohibit.
V2H1H2V1D1D2 is prohibit.
V2H2H1V1D1D2 is prohibit.
But
V1H1H2D1D2V2 is not prohibit.
V1H1D1H2D2V2 is not prohibit.
Where should I start?
or How to solve this problem?　
(Permutation)
I am sorry for my English Language
 A: If the "get rid of half repeatedly" argument is not satisfactory to you, we can approach directly via counting methods.


*

*Pick the locations occupied by the $V$'s.  $\binom{6}{2}$ choices

*Pick the locations occupied by the $D$'s.  $\binom{4}{2}$ choices

*Pick the locations occupied by the $H$'s.  $\binom{2}{2}$ choices


The real kicker here is the leftmost for each letter will be the one labeled with a 1 and the rightmost will be the one labeled with a two.
Apply multiplication principle to get the total number of arrangements is $\binom{6}{2}\binom{4}{2}\binom{2}{2}=15\cdot 6\cdot 1 = 90$
In other words, your question is in essence the same as the question of how many arrangements of the letters in the word VVDDHH exist, which we know to be $\binom{6}{2,2,2}=90$
A: In half of the permutations $V_2$ is before $V_1$ and in the other half $V_1$ is before $V_2$. Therefore, get rid of half of the solutions, leaving us with $360$.
In half of the permutations $H_2$ is before $H_1$ and in the other half $H_1$ is before $H_2$. Therefore, get rid of half of the solutions, leaving us with $180$.
In half of the permutations $D_2$ is before $D_1$ and in the other half $D_1$ is before $D_2$. Therefore, get rid of half of the solutions, leaving us with $90$.
