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Multinomial Expansion

The coefficie...

Question

The coefficient of $x^{2} y^{3}$ in the expansion of $(1 - x + y)^{20}$ is

\frac{20!}{2! 3!}

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- \frac{20!}{2! 3!}

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\frac{20!}{5! 2! 3!}

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\frac{20!}{(15!) (3!) (2!)}

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Solution

The correct option is D $\frac{20!}{(15!) (3!) (2!)}$
The general term in the expansion of $(1 - x + y)^{20}$
$= \frac{20!}{k_{1}! k_{2}! k_{3}!} 1^{k_{1}} (- x)^{k_{2}} (y)^{k_{3}}$

Where, $k_{1} + k_{2} + k_{3} = 20$
For $x^{2} y^{3}$ , so
$k_{2} = 2, k_{3} = 3 \Rightarrow k_{1} = 15$

Hence, the coefficient of $x^{2} y^{3}$
$= \frac{20!}{(15!) (3!) (2!)}$

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Multinomial Expansion

Standard XII Mathematics

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The coefficient of x2y3 in the expansion of (1−x+y)20 is

The correct option is D 20!(15!)(3!)(2!)The general term in the expansion of (1−x+y)20 =20!k1!k2!k3!1k1(−x)k2(y)k3 Where, k1+k2+k3=20 For x2y3, so k2=2,k3=3⇒k1=15 Hence, the coefficient of x2y3 =20!(15!)(3!)(2!)

The coefficient of $x^{2} y^{3}$ in the expansion of $(1 - x + y)^{20}$ is