Number of ways to pick a selection of coins How many ways are there to pick a selection of coins from $1 worth of identical pennies, 1 dollar worth of identical nickels, and 1 dollar worth of identical dimes if you select a total of 16 coins?
I know I start off with C(16+3-1,16), but I can't seem to factor the fact that you can't have 16 dimes.  
 A: Here is one approach. Let $d$ represent the number of dimes you choose. This number is distinguished from the other two, because it can't go past $10$, but the other two can go up to $16$. Let $n$ represent the number of nickels. With $d$ dimes chosen, you can have $0$ through $16-d$ nickels. Then the number of pennies is determined as $16-d-n$.
$$\begin{align}
\sum_{d=0}^{10}\sum_{n=0}^{16-d}1&=\sum_{d=0}^{10}(17-d)\\
&=17\cdot11-\binom{11}{2}\\
&=132
\end{align}$$
Alternatively,
$$\begin{align}
\sum_{d=0}^{10}\sum_{n=0}^{16-d}1&=\sum_{d=0}^{10}(17-d)=17+16+\cdots+7\\
&=\sum_{d=1}^{17}d-\sum_{d=1}^6d\\
&=\binom{18}{2}-\binom{7}{2}\\
&=132
\end{align}$$
A: We can use Stars and Bars, getting $\binom{18}{16}$, but then we must subtract the forbidden combinations that involve $11$ to $16$ dimes.
With $11$ dimes there are $6$ possibilities ($0$ to $5$ nickels, the rest pennies). With $12$ dimes there are $5$ possibilities, and so on down to $1$ possibility for $16$ dimes. Thus the total is $\binom{18}{16}-(6+5+\cdots+1)$. 
