AP EAMCET · Maths · Binomial Theorem
The coefficient of \(x^{10}\) in the expansion of \(\left(1+x^2-x^3\right)^8\) is
- A \(506\)
- B \(496\)
- C \(486\)
- D \(476\)
Answer & Solution
Correct Answer
(D) \(476\)
Step-by-step Solution
Detailed explanation
General term: \( \frac{8!}{p!q!r!} (x^2)^q (-x^3)^r \) where \( p+q+r=8 \) and \( 2q+3r=10 \). If \( r=0: 2q=10 \implies q=5 \). Then \( p=8-5-0=3 \). Coeff: \( \frac{8!}{3!5!0!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56 \). If \( r=2: 2q=4 \implies q=2 \). Then…
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