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 $\binom{6}{2,2,2}=90$