Q47 / 110 · VITEEE 2024 · Mathematics

Coefficient of $x^3$ in the expansion of $\left(x^2-x+1\right)^{10}\left(x^2+1\right)^{15}$ is equal to

  • Correct answer
    A
    -360
  • B
    -240
  • C
    -180
  • D
    60

Solution

$\text { } \begin{aligned} &\left(x^2-\right.x+1)^{10}\left(x^2+1\right)^{15} \\ & \quad=\left(1+{ }^{10} C_1\left(x^2-x\right)+{ }^{10} C_2\left(x^2-x\right)^2+\ldots\right) \\ & \quad\left(1+{ }^{15} C_1 x^2+{ }^{15} C_2\left(x^2\right)^2+\ldots\right) \end{aligned}$ We get, the coefficient of $x^3$. $\begin{aligned} & =-2 \cdot{ }^{10} C_2 \cdot{ }^{15} C_0-{ }^{10} C_1 \cdot{ }^{15} C_1-{ }^{10} C_3 \\ & =-2 \times \frac{10 \times 9}{2} \times 1-10 \times 15-\frac{10 \times 9 \times 8}{3 \times 2} \\ & =-360 \end{aligned}$

Hence, the correct answer is option: 'A'.

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