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

The **binomial theorem** states that for any expansion of the form \( (a + b)^n \), we have:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In this case:
- \( a = 3x^2 \)
- \( b = y \)
- \( n = 5 \)

So, the general term in the expansion is given by:

\[
\binom{5}{k} (3x^2)^{5-k} y^k
\]

Simplifying this, we get:

\[
\binom{5}{k} 3^{5-k} x^{2(5-k)} y^k
\]

We need to find the term where \( x^6 y^2 \) appears. This means:
- The power of \( x \) is 6, so \( 2(5-k) = 6 \), which gives \( k = 2 \).
- The power of \( y \) is 2, which matches \( k = 2 \).

Now, substitute \( k = 2 \) into the general term:

\[
\binom{5}{2} 3^{5-2} x^{2(5-2)} y^2 = \binom{5}{2} 3^3 x^6 y^2
\]

Now calculate:

\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = 10, \quad 3^3 = 27
\]

So, the coefficient of \( x^6 y^2 \) is:

\[
10 \times 27 = 270
\]

Thus, the coefficient of \( x^6 y^2 \) in the expansion of \( (3x^2 + y)^5 \) is **270**.

### Part (b)
We are asked to interpret the expression:

\[
[x_1 x_2 x_3^2] (x_1 + 2x_2 + 3x_3)^4
\]

This refers to a **multinomial expansion**. The multinomial expansion of \( (x_1 + 2x_2 + 3x_3)^4 \) can be written as:

\[
(x_1 + 2x_2 + 3x_3)^4 = \sum_{i_1 + i_2 + i_3 = 4} \binom{4}{i_1, i_2, i_3} (x_1)^{i_1} (2x_2)^{i_2} (3x_3)^{i_3}
\]

The notation \( [x_1 x_2 x_3^2] \) means we are looking for the term where \( x_1^1 x_2^1 x_3^2 \) appears. This corresponds to:
- \( i_1 = 1 \), \( i_2 = 1 \), and \( i_3 = 2 \).

For these values, the multinomial coefficient is:

\[
\binom{4}{1, 1, 2} = \frac{4!}{1! 1! 2!} = 12
\]

Now, the corresponding term is:

\[
12 \cdot (x_1)^1 \cdot (2x_2)^1 \cdot (3x_3)^2 = 12 \cdot 2 \cdot 9 \cdot x_1 x_2 x_3^2
\]

Thus, the coefficient of \( x_1 x_2 x_3^2 \) is:

\[
12 \times 2 \times 9 = 216
\]

The interpretation of this expression is that it represents the number of ways to select certain objects, accounting for different "weights" or counts of selections from each group, similar to a combinatorial counting problem.

---

Would you like further details or explanations? Here are 5 related questions that expand on this information:

1. How does the multinomial theorem generalize the binomial theorem?
2. Can the binomial coefficient \( \binom{n}{k} \) be used for other types of expansions?
3. How do we determine the total number of terms in a multinomial expansion?
4. What other types of problems can be solved using multinomial expansions?
5. How does the combinatorial interpretation of the binomial and multinomial theorems connect to probability theory?

**Tip:** Always check the powers of variables carefully when applying the binomial theorem to avoid mismatches.

Using the binomial theorem, determine the coefficient of x^6y^2 in the expansion of (3x^2 + y)^5. Interpret [x1x2x3^2](x1 + 2x2 + 3x3)^4 as the solution to some counting problem.

Learn how to apply the binomial theorem to find the coefficient of x^6y^2 in the expansion of (3x^2 + y)^5. This problem involves key concepts in algebra, such as binomial and multinomial expansions, and is suitable for university-level students studying advanced mathematics.

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