The coefficient of a term in the binomial expansion (a - b)^n can be found using the binomial theorem, which states that the coefficient is nCk * a^(n-k) * b^k, where nCk is "n choose k", a combination which can be calculated as n! / [k!(n-k)!].

In this case, we want the 4th term of the expansion of (x - 3y)^5, which means n = 5 and k = 3 (since the first term is k = 0).

So, the coefficient is 5C3 * x^(5-3) * (-3y)^3 = 10 * x^2 * (-27y^3) = -270x^2y^3.

Therefore, the coefficient of the 4th term (x^2y^3) in the expansion of (x - 3y)^5 is -270. So, the correct answer is c. -270.

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The coefficient of a term in the binomial expansion (a - b)^n can be found using the binomial theorem, which states that the coefficient is nCk * a^(n-k) * b^k, where nCk is "n choose k", a combination which can be calculated as n! / [k!(n-k)!].

In this case, we want the 4th term of the expansion of (x - 3y)^5, which means n = 5 and k = 3 (since the first term is k = 0).

So, the coefficient is 5C3 * x^(5-3) * (-3y)^3 = 10 * x^2 * (-27y^3) = -270x^2y^3.

Therefore, the coefficient of the 4th term (x^2y^3) in the expansion of (x - 3y)^5 is -270. So, the correct answer is c. -270.

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The coefficient of a term in the binomial expansion (a - b)^n can be found using the binomial theorem, which states that the coefficient is nCk * a^(n-k) * b^k, where nCk is "n choose k", a combination which can be calculated as n! / [k!(n-k)!].

In this case, we want the 4th term of the expansion of (x - 3y)^5, which means n = 5 and k = 3 (since the first term is k = 0).

So, the coefficient is 5C3 * x^(5-3) * (-3y)^3 = 10 * x^2 * (-27y^3) = -270x^2y^3.

Therefore, the coefficient of the 4th term (x^2y^3) in the expansion of (x - 3y)^5 is -270. So, the correct answer is c. -270.

What is the coefficient of the 4th term (𝑥2𝑦3) in the expansion of (𝑥 − 3𝑦)5a. 270b. 190c. −270d. 720

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