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# Find the coefficient of a⁵b⁷ in (a - 2b)¹²

**Solution:**

It is known that (r + 1)^{th} term, (Tᵣ ₊ ₁) in the binomial expansion of (a + b)^{n} is given by Tᵣ ₊ ₁ = ⁿCᵣ aⁿ ⁻ ʳ bʳ.

Assuming a⁵b⁷ occurs in the expansion of (a - 2b)¹², we obtain

Tᵣ ₊ ₁ = ¹²Cᵣ(a)¹² ⁻ ʳ (- 2b)ʳ

= ¹²Cᵣ(- 2)ʳ (a)¹² ⁻ ʳ (b)ʳ

Comparing the exponents of x in a⁵b⁷ in (Tᵣ ₊ ₁) we obtain

12 - r = 5 (or) r = 7

From each of these, we get r = 7

Thus, the coefficient of a⁵b⁷ is ¹²C₇(- 2)⁷.

¹²C₇(- 2)⁷ = [12!/(7! 5!)] x (- 2)⁷

= [12 x 11 x 10 x 9 x 8 x (7!)]/[(7!) (5!)] x (- 128)

= - (792) x (128)

= - 101376

NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 2

## Find the coefficient of a⁵b⁷ in (a - 2b)¹²

**Summary:**

Using the binomial theorem, the coefficient of a⁵b⁷ in (a - 2b)¹² is to be evaluated. We have found that it is equal to - 101376

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