In the expansion of (2x + 3y)^4, what is the coefficient of x^2y^2?

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Q: In the expansion of (2x + 3y)^4, what is the coefficient of x^2y^2?

Solution: The coefficient is C(4,2) * (2)^2 * (3)^2 = 6 * 4 * 9 = 216.

Steps: 12

Step 1: Identify the expression to expand, which is (2x + 3y)^4.
Step 2: Recognize that we need to find the coefficient of the term x^2y^2 in the expansion.
Step 3: Use the Binomial Theorem, which states that (a + b)^n = Σ [C(n, k) * a^(n-k) * b^k] for k = 0 to n.
Step 4: In our case, a = 2x, b = 3y, and n = 4.
Step 5: We want the term where x has an exponent of 2 and y has an exponent of 2, which means we need k = 2 (since y is raised to the power of 2).
Step 6: Calculate n - k, which is 4 - 2 = 2. This means we need the coefficient for C(4, 2).
Step 7: Calculate C(4, 2), which is the number of ways to choose 2 items from 4. C(4, 2) = 4! / (2! * (4-2)!) = 6.
Step 8: Calculate (2)^2, which is 4.
Step 9: Calculate (3)^2, which is 9.
Step 10: Multiply the results: Coefficient = C(4, 2) * (2)^2 * (3)^2 = 6 * 4 * 9.
Step 11: Perform the multiplication: 6 * 4 = 24, then 24 * 9 = 216.
Step 12: Conclude that the coefficient of x^2y^2 in the expansion of (2x + 3y)^4 is 216.