What is the probability of rolling 3 different pairs with seven dice? for example one outcome can be this :
(1,1,2,2,3,3,4)
My answer is different compared to the book
This is how i solve it: 
$$\frac{\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{6}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}}{6^7} $$
The book solves it like this:
$$\frac{\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{6}{3}\binom{3}{1}}{6^7}=0.135  $$
am i wrong?
isn't this outcome :(1,1,2,2,3,3,4) different compared to this :(2,2,1,1,3,3,4) ?

Depending on whether a triple covers a double, we can include the red term:
$$
\color{#C00}{\underbrace{\overbrace{\ \ \ \binom{6}{2}\ \ \ }^\text{$2$ pair}\overbrace{\ \ \ \binom{4}{1}\ \ \ }^\text{$1$ triple}\overbrace{\vphantom{\binom{6}{2}}\ \frac{7!}{2!2!3!}\ }^\text{rearrangements}}_{\sim4.5\%\text{ of }6^7}+}\underbrace{\overbrace{\ \ \ \binom{6}{3}\ \ \ }^\text{$3$ pair}\overbrace{\ \ \ \binom{3}{1}\ \ \ }^\text{$1$ singleton}\overbrace{\vphantom{\binom{6}{2}}\frac{7!}{2!2!2!1!}}^\text{$\ \ $rearrangements}}_{\sim13.5\%\text{ of }6^7}
$$
Note that $\frac{7!}{2!2!2!1!}=\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}$ and $\frac{7!}{2!2!3!}=\binom{7}{2}\binom{5}{2}\binom{3}{3}$, so the answer, if not accepting a triple, would be
$$
\left.\binom{6}{3}\binom{3}{1}\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}\middle/6^7\right.\approx13.5\%
$$
which matches the book answer.
