First slide

Binomial theorem for positive integral Index

Question

$The coefficient of x^{3} y^{4} z^{2} in the expansion of$ $(2 x - 3 y + 4 z)^{9} is$

Moderate

A

12063600
B

13063680
C

11063600
D

None of these above

Solution

The general term in the expansion of $(2 x - 3 y + 4 z)^{9}$
$\begin{aligned} = \frac{9!}{α_{1}! α_{2}! α_{3}!} \cdot (2 x)^{α_{1}} (- 3 y)^{α_{2}} (4 z)^{α_{3}} \\ = \frac{9!}{α_{1}! α_{2}! α_{3}!} \cdot 2^{α_{1}} (- 3)^{α_{2}} 4^{α_{3}} \cdot x^{α_{1}} y^{α_{2}} z^{α_{3}} \end{aligned}$
$∴ Coefficient of x^{3} y^{4} z^{2}$
$= \frac{9!}{3! 4! 2!} \cdot 2^{3} (- 3)^{4} \cdot 4^{2}$ $[Taking α_{1} = 3, α_{2} = 4, α_{3} = 2]$
$\begin{aligned} = 1260 \times 10368 \\ = 13063680 \end{aligned}$

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