We are tasked with determining how many ways Alexandra can take 33 additional questions from the 9 other distinguishable grading TAs, with the condition that at least 4 of the TAs give her exactly 7 questions each to grade.

Step 1: Define the problem clearly

Let $x_i$ represent the number of questions Alexandra takes from TA $i$, where $i = 1, 2, \dots, 9$. Therefore, we have the equation:

$x_1 + x_2 + \cdots + x_9 = 33$

with the restriction that $0 \leq x_i \leq 7$ for all $i$, and at least 4 of the $x_i$'s must be equal to 7.

Step 2: Handle the restriction of at least 4 TAs giving 7 questions

Let us begin by selecting 4 TAs who will give Alexandra exactly 7 questions each. There are $\binom{9}{4}$ ways to choose these 4 TAs. For each of these selected TAs, we set $x_i = 7$.

This means that for these 4 TAs, $4 \times 7 = 28$ questions are already accounted for. Thus, the remaining $33 - 28 = 5$ questions need to be distributed among the remaining $9 - 4 = 5$ TAs. For these remaining TAs, each $x_i$ can take values between 0 and 7.

Step 3: Counting the number of ways to distribute the remaining questions

We now need to distribute 5 questions among 5 TAs, with each TA allowed to receive between 0 and 6 questions (because they cannot receive more than 7 total, and we’ve already assigned at least 4 TAs the maximum of 7 questions). This is a bounded stars and bars problem.

We will solve this by transforming the variables. Let $y_i = 7 - x_i$, where $y_i \geq 1$ for the remaining TAs, since each of these TAs can receive up to 6 questions. Now, the equation becomes:

$y_1 + y_2 + \cdots + y_5 = 33 - 28 = 5$

We need to count the number of solutions to this equation,