determine the number of r-combinations of the multiset Determine the number of r-combinations of the multiset $\{1  \cdot a_1, \infty.  \cdot a_2, ... , \infty \cdot a_k \}$.
The answer in the back of the book is $\binom{r+k-2}{k-2}+\binom{r+k-3}{k-2}$.
I see where $\binom{r+k-2}{k-2}$ comes from because it is the number of r-combinations from the multiset that does not include $a_1$. I don't understand where the $\binom{r+k-3}{k-2}$ comes from though. 

Let $x_j$ be the number of occurrences of $a_j$, where $1 \leq j \leq k$.  Then we need to find the number of solutions of the equation 
$$x_1 + x_2 + x_3 + \ldots + x_k = r \tag{1}$$
in the nonnegative integers subject to the restriction that $x_1 \leq 1$.  
If $x_1 = 0$, equation 1 reduces to 
$$x_2 + x_3 + \ldots + x_k = r \tag{2}$$
Since there are $k - 1$ terms in the sum, a particular solution of equation 2 corresponds to the placement of $k - 2$ addition signs in a row of $r$ ones.  The number of such solutions is 
$$\binom{r + k - 2}{k - 2}$$
since we must choose which $k - 2$ of the $r + k - 2$ positions for $r$ ones and $k - 2$ addition signs will be filled with addition signs.
If $x_1 = 1$, equation 1 reduces to 
$$x_2 + x_3 + \ldots + x_k = r - 1 \tag{3}$$
In this case, a particular solution of equation 3 corresponds to the placement of $k - 2$ addition signs in a row of $r - 1$ ones.  The number of such solutions is 
$$\binom{r - 1 + k - 2}{k - 2} = \binom{r + k - 3}{k - 2}$$
since we must choose which $k - 2$ of the $r - 1 + k - 2 = r + k - 3$ positions for $r - 1$ ones and $k - 2$ addition signs will be filled with addition signs.
Since the two cases are mutually exclusive and exhaustive, the number of solutions of equation 1 is found by adding the results for equations 2 and 3.  
