Coefficient of Power of x in Product of Binomial Expansions

To determine the coefficient of $x^{11}$ in the polynomial expansion of: $P(x) = (1 + x^2)^4 (1 + x^3)^7 (1 + x^4)^{12}$

we first write each factor using the binomial theorem: $(1 + x^2)^4 = \sum_{a=0}^{4} \binom{4}{a} x^{2a}$ $(1 + x^3)^7 = \sum_{b=0}^{7} \binom{7}{b} x^{3b}$ $(1 + x^4)^{12} = \sum_{c=0}^{12} \binom{12}{c} x^{4c}$

The general term in the product is: $\binom{4}{a} \binom{7}{b} \binom{12}{c} x^{2a + 3b + 4c}$

To find the coefficient of $x^{11}$, we need to find all non-negative integers $a, b, c$ such that: $2a + 3b + 4c = 11$ subject to the constraints: $0 \le a \le 4, \quad 0 \le b \le 7, \quad 0 \le c \le 12$

Since $2a$, $4c$, and $11$ are even, even, and odd respectively, $3b$ must be odd. Hence, $b$ must be an odd integer.

Given $3b \le 11$, the possible values for $b$ are $1$ and $3$.

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Case 1: $b = 1$

Substitute $b = 1$ into the equation: $2a + 3(1) + 4c = 11 \implies 2a + 4c = 8 \implies a + 2c = 4$

The possible non-negative integer solutions $(a, c)$ are:

1. For $c = 0 \implies a = 4$: $\text{Triplet: } (a, b, c) = (4, 1, 0)$ $\text{Term} = \binom{4}{4} \binom{7}{1} \binom{12}{0} = 1 \times 7 \times 1 = 7$

2. For $c = 1 \implies a = 2$: $\text{Triplet: } (a, b, c) = (2, 1, 1)$ $\text{Term} = \binom{4}{2} \binom{7}{1} \binom{12}{1} = 6 \times 7 \times 12 = 504$

3. For $c = 2 \implies a = 0$: $\text{Triplet: } (a, b, c) = (0, 1, 2)$ $\text{Term} = \binom{4}{0} \binom{7}{1} \binom{12}{2} = 1 \times 7 \times \frac{12 \times 11}{2} = 7 \times 66 = 462$

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Case 2: $b = 3$

Substitute $b = 3$ into the equation: $2a + 3(3) + 4c = 11 \implies 2a + 4c = 2 \implies a + 2c = 1$

The only non-negative integer solution $(a, c)$ is:

1. For $c = 0 \implies a = 1$: $\text{Triplet: } (a, b, c) = (1, 3, 0)$ $\text{Term} = \binom{4}{1} \binom{7}{3} \binom{12}{0} = 4 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 1 = 4 \times 35 \times 1 = 140$

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Sum of all coefficients:

$\text{Coefficient of } x^{11} = 7 + 504 + 462 + 140 = 1113$

Thus, the correct option is (C).