A) \[\frac{9!}{3!6!}\]
B) \[\frac{9!}{4!5!}\]
C) \[\frac{9!}{3!5!}\]
D) \[\frac{9!}{3!4!}\]
Correct Answer: D
Solution :
[d] \[{{(1+a-b+c)}^{9}}\] \[=\sum{\frac{9!}{{{x}_{1}}!{{x}_{2}}!{{x}_{3}}!{{x}_{4}}!}.{{(1)}^{x1}}{{(a)}^{x2}}{{(-b)}^{x3}}{{(x)}^{x4}}}\] \[\Rightarrow \] Coefficient of \[{{a}^{3}}{{b}^{4}}c=\frac{9!}{1!3!4!1!}=\frac{9!}{3!4!}\]
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