Coefficient of $x^{11}$ in the expansion of $\left(1+x^{2}\right)^{4}\left(1+x^{3}\right)^{7}\left(1+x^{4}\right)^{12}$ is

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Binomial Theorem

Coefficient of

x^{11}

in the expansion of

(1 + x^{2})^{4} (1 + x^{3})^{7} (1 + x^{4})^{12}

Answer: (C)

To find the coefficient of

x^{11}

, we need to find all combinations of powers from the three factors that sum to 11.

The possible combinations are:

$x^{0}$ from first, $x^{3}$ from second, $x^{8}$ from third: $(0 4) \times (1 7) \times (2 12) = 1 \times 7 \times 66 = 462$
$x^{8}$ from first, $x^{3}$ from second, $x^{0}$ from third: $(4 4) \times (1 7) \times (0 12) = 1 \times 7 \times 1 = 7$
$x^{2}$ from first, $x^{9}$ from second, $x^{0}$ from third: $(1 4) \times (3 7) \times (0 12) = 4 \times 35 \times 1 = 140$
$x^{4}$ from first, $x^{3}$ from second, $x^{4}$ from third: $(2 4) \times (1 7) \times (1 12) = 6 \times 7 \times 12 = 504$

Total coefficient =

462 + 7 + 140 + 504 = 1113

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option