The coefficient of x^4 in the expansion of is (where

JEE MainMathematicsToughBinomial Theorem

The coefficient of x^4 in the expansion of
$\left( 1 + x + x^{2} + x^{3} \right)^{11}$ is
(where $(n\ r\ ) = \ ^{n}C_{r})$

Show answer and solution

1 + x + x^{2} + x^{3} = (1 + x) + x^{2}(1 + x) = (1 + x)\left( 1 + x^{2} \right)

\left( 1 + x + x^{2} + x^{3} \right)^{11} = (1 + x)^{11}\left( 1 + x^{2} \right)^{11}

We want the coefficient of x^4

\left( 1 + \ ^{11}C_{1}.x + \ ^{11}C_{2}.x^{2} + \ ^{11}C_{3}.x^{3} + \ ^{11}C_{4}.x^{4} + \ldots \right) \times \left( 1 + \ ^{11}C_{1}.x^{2} + \ ^{11}C_{2}.{\ x}^{4} + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ldots. \right)

\therefore

The coefficient of x^4 is

\ ^{11}C_{2} + \ ^{11}C_{2}\ ^{11}C_{1} + \ ^{11}C_{4} = 55 + 605 + 330 = 990

Source: Grass-G / M-08 Mathematics Paper 23 October.docx

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