The coefficient of $a^{3} b^{4} c$ in the expansion of $(1 + a - b + c)^{9}$ is equal to

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Solution

The correct option is D
$(1 + a - b + c)^{9} = \sum \frac{9!}{x_{1}! x_{2}! x_{3}! x_{4}} . (1)^{x_{2}} (a)^{x_{2}} (- b)^{x_{3}} (C)^{x_{4}} \Rightarrow C o e f f i c i e n t o f a^{3} b^{4} c = \frac{9!}{1! 3! 4! 1!} = \frac{9!}{3! 4!}$

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Similar questions

If
A = {1, 2, 3, 4}

B = {3, 4, 5, 6, 7}

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D = {7, 8, 9, 10}

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The coefficient of a3b4c in the expansion of (1+a–b+c)9 is equal to

The correct option is D (1+a−b+c)9=∑9!x1!x2!x3!x4.(1)x2(a)x2(−b)x3(C)x4⇒Coefficient of a3b4c=9!1!3!4!1!=9!3!4!

The coefficient of $a^{3} b^{4} c$ in the expansion of $(1 + a - b + c)^{9}$ is equal to

The correct option is D
$(1 + a - b + c)^{9} = \sum \frac{9!}{x_{1}! x_{2}! x_{3}! x_{4}} . (1)^{x_{2}} (a)^{x_{2}} (- b)^{x_{3}} (C)^{x_{4}} \Rightarrow C o e f f i c i e n t o f a^{3} b^{4} c = \frac{9!}{1! 3! 4! 1!} = \frac{9!}{3! 4!}$