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?

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 
$$\binom{26}{6} - \binom{20}{6}$$
