Counting Multi-Sets for Donuts The problem is:
Suzy is selecting 20 donuts to bring to her club meeting. The donut store sells 7 varieties of donuts. Donuts of the same variety are all the same. There is a large supply of each variety of donuts, except for jelly donuts. There are only 5 jelly donuts available. How many ways are there for Suzy to select the donuts?
I was thinking that the answer was $5{20 + 6 - 1 \choose 6-1}$, since there is not an unlimited supply of jelly donuts. Would that logic be correct?
 A: Suzy can choose $j\in[0..5]$ jelly donuts and then $20-j$ donuts of the other six kinds. Therefore there are
$$\sum_{j=0}^5 {20-j+5\choose 5}=191\,470$$
ways to chose the donuts. The number ($265\,650$) proposed by you is wrong on two accounts: (i) There are six possible choices for $j$. (ii) When the $j$ jelly donuts have been chosen there are no longer $20$ donuts to be bought.
A: Christian Blatter has explained how to correct your approach and why your approach was incorrect.  Here is another method.
Let $x_j$ denote the number of jelly donuts.  Let $x_k$, $1 \leq k \leq 6$, be the number of donuts of type $k$ that Suzy purchases.  Then 
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_j = 20 \tag{1}$$
is an equation in the nonnegative integers.  If we temporarily ignore the restriction that $x_j \leq 5$, a particular solution of equation 1 corresponds to the placement of six addition signs in a row of $20$ ones.  For instance,
$$1 1 1 + 1 1 + 1 1 1 1 + 1 1 + + 1 1 1 1 1 1 + 1 1 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 2$, $x_3 = 4$, $x_4 = 2$, $x_5 = 0$, $x_6 = 6$, and $x_j = 3$.  There are 
$$\binom{20 + 6}{6} = \binom{26}{6}$$
such solutions since we must choose which $6$ of the $26$ positions required for $20$ ones and $6$ addition signs will be filled with addition signs.  
From these, we must subtract the number of cases in which the condition $x_j \leq 5$ is violated.  Suppose $x_j > 5$. Then $x_j' = x_j - 6$ is a nonnegative integer.  Substituting $x_j' + 6$ for $x_j$ in equation 1 yields
\begin{align*}
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_j' + 6 & = 20\\
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_j' & = 14 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with 
$$\binom{14 + 6}{6} = \binom{20}{6}$$
solutions.  Since these are the cases that violate the restriction that $x_j \leq 5$, we conclude that the number of ways that Suzy can purchase $20$ donuts at the store without buying more than five jelly donuts is 
$$\binom{26}{6} - \binom{20}{6}$$
