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) ?
 A: Here is a justification of the book's answer:  
Choose the three values for the pairs from the six possible values on the dice in $\binom{6}{3}$ ways.  Choose the value of the singleton from the remaining three values in $\binom{3}{1}$ ways.  For the three pairs, there are $\binom{7}{2}$ ways for the smallest value to appear on two of the seven dice, $\binom{5}{2}$ ways for the middle value to appear on two of the other five dice, and $\binom{3}{2}$ ways for the largest value to appear on two of the other three dice.  The remaining die shows the singleton value.  Hence, there are 
$$\binom{6}{3}\binom{3}{1}\binom{7}{2}\binom{5}{2}\binom{3}{2}$$
favorable outcomes.
In your attempt, you count each favorable outcome six times, once for each way you designate one of the three pairs as your first pair, another pair as your second pair, and the remaining pair as your third pair.  The order in which the pairs are selected does not matter.  Notice that 
$$6\binom{6}{3}\binom{3}{1}\binom{7}{2}\binom{5}{2}\binom{3}{2} = \binom{6}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{7}{2}\binom{5}{2}\binom{3}{2}$$
A: 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.
