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Observe that 10!/1!9! = ¹⁰C₁ , 10!/3!7! = ¹⁰C₃ , 10!/5!5! = ¹⁰C₅ which are alternate coefficients in the expansion of (1+x)ⁿ which can be found out by :

(1+x)¹⁰ - (1-x)¹⁰ = 2[¹⁰C₁x¹ + ¹⁰C₃x³ + ¹⁰C₅x⁵ + ¹⁰C₇x⁷ + ¹⁰C₉x⁹]

Putting x = 1

2¹⁰ = 2[¹⁰C₁ + ¹⁰C₃ + ¹⁰C₅ + ¹⁰C₇ + ¹⁰C₉]

Since ¹⁰C₁ = ¹⁰C₉ , ¹⁰C₃ = ¹⁰C₇

Hence 2⁹ = [2¹⁰C₁ + 2¹⁰C₃ + ¹⁰C₅] = [2(10!)/(1!9!) + 2(10!)/(3!7!) + 10!/(5!5!)]

Hence [2/(1!9!) + 2/(3!7!) + 1/(5!5!)] = 2⁹/10! = 2^a/b!

Hence a = 9, b = 10

(x-2)² = (5i)² gives x²-4x+4 = -25

x² = 4x - 29

x³ = x(x²) = 4x² - 29x = 4(4x - 29) - 29x = -13x - 108

Hence x³ - 5x² + 33x - 19 = (-13x - 108) - 5(4x - 29) + 33x - 19 = 10 = b

Hence answer is c) b.

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