Step-by-step Derivations

1. Identify the General Term

The given expression is $P(x) = {\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$.

We can expand each factor using the binomial theorem:

• ${\left( {1 + {x^2}} \right)^4} = \sum\limits_{i = 0}^4 {\binom{4}{i}{{\left( {{x^2}} \right)}^i}} = \sum\limits_{i = 0}^4 {\binom{4}{i}{x^{2i}}}$
• ${\left( {1 + {x^3}} \right)^7} = \sum\limits_{j = 0}^7 {\binom{7}{j}{{\left( {{x^3}} \right)}^j}} = \sum\limits_{j = 0}^7 {\binom{7}{j}{x^{3j}}}$
• ${\left( {1 + {x^4}} \right)^{12}} = \sum\limits_{k = 0}^{12} {\binom{12}{k}{{\left( {{x^4}} \right)}^k}} = \sum\limits_{k = 0}^{12} {\binom{12}{k}{x^{4k}}}$

A general term in the expansion of the product $P(x)$ is the product of the general terms of each factor: $\left( \binom{4}{i}x^{2i} \right) \left( \binom{7}{j}x^{3j} \right) \left( \binom{12}{k}x^{4k} \right) = \binom{4}{i}\binom{7}{j}\binom{12}{k} x^{2i+3j+4k}$ where the indices $i, j, k$ are integers satisfying: $0 \le i \le 4$, $0 \le j \le 7$, $0 \le k \le 12$.

2. Set up the Equation for the Exponent

To find the coefficient of ${x^{11}}$, we need the exponent of $x$ to be 11. This gives us the linear Diophantine equation: $2i + 3j + 4k = 11$

3. Find All Integer Solutions

We need to find all non-negative integer solutions $(i, j, k)$ to this equation that satisfy the given constraints on $i, j,$ and $k$. We can systematically find the solutions by iterating through the possible values of $k$, as it has the largest coefficient.

Since $4k \le 11$, the possible values for $k$ are $0, 1, 2$.

Case 1: $k = 0$ The equation becomes $2i + 3j = 11$.

• If $j=0$, $2i=11$ (no integer solution for $i$).
• If $j=1$, $2i=8 \implies i=4$. This gives the triplet $(4, 1, 0)$. This is a valid solution as $i=4 \le 4$ and $j=1 \le 7$.
• If $j=2$, $2i=5$ (no integer solution for $i$).
• If $j=3$, $2i=2 \implies i=1$. This gives the triplet $(1, 3, 0)$. This is a valid solution as $i=1 \le 4$ and $j=3 \le 7$.
• If $j \ge 4$, $3j \ge 12 > 11$, so no solution with non-negative $i$.

Case 2: $k = 1$ The equation becomes $2i + 3j + 4(1) = 11$, which simplifies to $2i + 3j = 7$.

• If $j=0$, $2i=7$ (no integer solution for $i$).
• If $j=1$, $2i=4 \implies i=2$. This gives the triplet $(2, 1, 1)$. This is a valid solution as $i=2 \le 4$ and $j=1 \le 7$.
• If $j=2$, $2i=1$ (no integer solution for $i$).
• If $j \ge 3$, $3j \ge 9 > 7$, so no solution with non-negative $i$.

Case 3: $k = 2$ The equation becomes $2i + 3j + 4(2) = 11$, which simplifies to $2i + 3j = 3$.

• If $j=0$, $2i=3$ (no integer solution for $i$).
• If $j=1$, $2i=0 \implies i=0$. This gives the triplet $(0, 1, 2)$. This is a valid solution as $i=0 \le 4$ and $j=1 \le 7$.
• If $j \ge 2$, $3j \ge 6 > 3$, so no solution with non-negative $i$.

Case 4: $k \ge 3$ If $k=3$, $4k=12 > 11$, so there are no solutions for $k \ge 3$ with non-negative $i$ and $j$.

So, the valid triplets $(i, j, k)$ are: $(4, 1, 0)$, $(1, 3, 0)$, $(2, 1, 1)$, and $(0, 1, 2)$.

4. Calculate and Sum the Coefficients

The total coefficient of $x^{11}$ is the sum of the coefficients for each valid triplet:

• For $(4, 1, 0)$: Coefficient is $\binom{4}{4}\binom{7}{1}\binom{12}{0} = 1 \times 7 \times 1 = 7$.
• For $(1, 3, 0)$: Coefficient is $\binom{4}{1}\binom{7}{3}\binom{12}{0} = 4 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 1 = 4 \times 35 \times 1 = 140$.
• For $(2, 1, 1)$: Coefficient is $\binom{4}{2}\binom{7}{1}\binom{12}{1} = \frac{4 \times 3}{2 \times 1} \times 7 \times 12 = 6 \times 7 \times 12 = 504$.
• For $(0, 1, 2)$: Coefficient is $\binom{4}{0}\binom{7}{1}\binom{12}{2} = 1 \times 7 \times \frac{12 \times 11}{2 \times 1} = 7 \times 66 = 462$.

Total coefficient = $7 + 140 + 504 + 462 = 1113$.

Conclusion

The coefficient of ${x^{11}}$ in the given expansion is 1113. This corresponds to option C.