Alexandra has finished grading the questions she's been assigned and wants to help the 9 other distinguishable grading TAs with the indistinguishable questions they are still grading. How many ways can Alexandra take 33 additional questions from the other grading TAs, such that at least 4 of the other grading TAs give her 7 questions each to grade.

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,

Explore how to solve a combinatorics problem involving distributing 33 questions among 9 grading TAs, where at least 4 must give exactly 7 questions. The solution uses the Stars and Bars method and binomial coefficients, making it suitable for advanced high school or undergraduate students.

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