https://file.aiquickdraw.com/mathbot/file/1726278505308f83kgzbz.jpg

Let's work through the questions from the image.

### Part (a)
We are asked to find the coefficient of \(x^6y^2\) in the expansion of \((3x^2 + y)^5\) using the **binomial theorem**.

The binomial theorem states that:
\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]
For the expansion of \((3x^2 + y)^5\):
- \(a = 3x^2\),
- \(b = y\),
- \(n = 5\).

To find the coefficient of \(x^6y^2\), we need to choose terms such that the powers of \(x\) and \(y\) match \(x^6y^2\). Notice:
- The term \(x^6\) comes from \((3x^2)^3 = 27x^6\),
- The term \(y^2\) comes from \(y^2\).

This means we need the term in the binomial expansion where \(k = 2\) (so that we get \(y^2\)). The general term in the expansion is:
\[
\binom{5}{k} (3x^2)^{5-k} y^k
\]
For \(k = 2\), this becomes:
\[
\binom{5}{2} (3x^2)^{3} y^2 = \binom{5}{2} \cdot 27x^6 \cdot y^2
\]
Now, calculate the binomial coefficient:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
Thus, the coefficient of \(x^6y^2\) is:
\[
10 \cdot 27 = 270
\]
So, the coefficient of \(x^6y^2\) in the expansion of \((3x^2 + y)^5\) is **270**.

### Part (b)
We are asked to interpret:
\[
[x_1x_2x_3^2](x_1 + 2x_2 + 3x_3)^4
\]
This is a problem in combinatorics. We are looking for the coefficient of the term \(x_1x_2x_3^2\) in the expansion of \((x_1 + 2x_2 + 3x_3)^4\).

This expression can be understood as counting how many ways the variables \(x_1\), \(x_2\), and \(x_3\) can be selected to form the term \(x_1x_2x_3^2\). The expansion represents a multinomial, and the task is to determine the number of ways to obtain the specific combination of powers \(x_1^1x_2^1x_3^2\) when distributing four total factors from \((x_1 + 2x_2 + 3x_3)^4\).

For each selection:
- Picking \(x_1\) from the first term contributes 1 to the total count,
- Picking \(x_2\) from the second term contributes 2 to the total count,
- Picking \(x_3\) from the third term contributes 3 to the total count.

To form the desired term \(x_1x_2x_3^2\), the multinomial coefficient is given by:
\[
\text{Coefficient of } x_1x_2x_3^2 = \frac{4!}{1!1!2!} \cdot (1^1)(2^1)(3^2)
\]
We will calculate this:

\[
\text{Coefficient} = \frac{4!}{1!1!2!} \cdot 1 \cdot 2 \cdot 9 = \frac{24}{2} \cdot 18 = 12 \cdot 18 = 216
\]

So, the coefficient of \(x_1x_2x_3^2\) is **216**, which can be interpreted as the number of ways to select the variables \(x_1\), \(x_2\), and \(x_3^2\) in the given multinomial expansion.

---

Would you like further details or explanations on any part of the solution? Here are five related questions to explore the concepts:

1. How does the binomial theorem apply to non-linear terms like \(x^2\)?
2. What is the significance of binomial coefficients in expansions?
3. How do multinomial coefficients generalize the binomial theorem?
4. What combinatorial interpretation can be made for the term \(x^2y^3\) in a different expansion?
5. Can you compute the coefficient of \(x^4\) in \((2x + 3y)^6\)?

**Tip:** Understanding the binomial theorem helps simplify calculations in probability, combinatorics, and algebraic expansions.

Using the binomial theorem, determine the coefficient of x^6y^2 in the expansion of (3x^2 + y)^5. Verify your answer by actually computing the expansion. Interpret [x_1x_2x_3^2](x_1 + 2x_2 + 3x_3)^4 as the solution to some counting problem.

Explore how to determine the coefficient of x^6y^2 using the binomial theorem in the expansion of (3x^2 + y)^5. Learn about combinatorics and multinomial coefficients through the interpretation of [x_1x_2x_3^2](x_1 + 2x_2 + 3x_3)^4. This problem applies algebraic and combinatorial techniques suited for advanced mathematics students.

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