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Q. Coefficient of $x^4$ in $(1 + x + x^2 + x^3)^{11}$ is

Binomial Theorem

Solution:

We have, $(1 +x + x^2 + x^3)^{11} = (1 +x)^{11}\, (1 + x^2)^{11}$
$= (1 + \,{}^{11}C_1 x + \,{}^{11}C_2\, x^2 + \,{}^{11}C_3 \,x^3 + \,{}^{11}C_4\,x^4 + ...) \times$
$(1 + \,{}^{11}C_1\, x_2 + \,{}^{11}C_2\, (x^2)^2 + ...)$
$\therefore $ Coeffficient of $x^4 = \,{}^{11}C_2 + \,{}^{11}C_2 \times \,{}^{11}C_1 + \,{}^{11}C_4$
$= 55 + 605 + 330 = 990$