Counting Distinguishable Dominoes Pg. 19 of Counting: The Art of Enumerative Combinatorics, George E. Martin


*How many distinguishable dominoes are there, (each of the 2 ends of a domino has 0 to 6 dots carved on it)?


My answer is the following:
$\frac{{7+2-1}\choose{2}}{2!}$
I would like to know if this answer is correct and if not what is the correct answer and you was it derived?
 A: There are seven ways to fill an end of a domino.  Either both ends are the same or they are different.  There are $\binom{7}{1}$ ways to make both ends the same and $\binom{7}{2}$ ways to make the ends of a domino different.  Hence, the number of distinguishable dominoes is 
$$\binom{7}{1} + \binom{7}{2}$$
I assume you obtained your answer by finding the number of nonnegative integers solutions of the equation
$$x_0 + x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 2$$
where $x_i$ is the number of times $i$ dots appear at an end of the domino.  This equation has 
$$\binom{2 + 7 - 1}{7 - 1} = \binom{8}{6} = \binom{8}{2} = \binom{7}{1} + \binom{7}{2}$$
solutions.  Notice that each domino is only counted once with this method since $x_1 = 2$ means both ends of the domino have one dot and the solution $x_2 = 1$, $x_5 = 1$ means that one end of the domino has two dots while the other end has five dots.  We are not distinguishing which end of the domino has two dots and which end has five dots, so there is no need to divide by $2$, as you did.
