Animated Solution for Mathematics - Binomial Theorem: Coefficient of $x^{11}$ in the expansion of $(1+x^2{)}^4(1+x^3{)}^7(1+x^4{)}^{12}$ is

To find the coefficient of $x^{11}$ in the expansion of $(1+x^2{)}^4(1+x^3{)}^7(1+x^4{)}^{12}$, we treat the problem as a selection process from three distinct binomial sources.

Each source provides terms of the form: - Chest 1: ${}^4{C}_a(x^2{)}^a$ where $0\le a\le 4$ - Chest 2: ${}^7{C}_b(x^3{)}^b$ where $0\le b\le 7$ - Chest 3: ${}^{12}{C}_c(x^4{)}^c$ where $0\le c\le 12$

The general term of the product is given by the product of these selections:

We require the total exponent of $x$ to be 11. Thus, we must solve the following Diophantine equation under the given constraints:

We iterate through possible values of $c$ to find integer solutions for $a$ and $b$:

Case 1: $c=0$ The equation becomes $2a+3b=11$. - If $b=1$, then $2a=8\Rightarrow a=4$. Triplet: $(4,1,0)$. - If $b=3$, then $2a=2\Rightarrow a=1$. Triplet: $(1,3,0)$.

Case 2: $c=1$ The equation becomes $2a+3b=7$. - If $b=1$, then $2a=4\Rightarrow a=2$. Triplet: $(2,1,1)$.

Case 3: $c=2$ The equation becomes $2a+3b=3$. - If $b=1$, then $2a=0\Rightarrow a=0$. Triplet: $(0,1,2)$.

For $c\ge 3$, the term $4c\ge 12$, which exceeds our target of 11. No further solutions exist.

We now calculate the coefficient for each valid triplet $(a,b,c)$ using the product of combinations:

1. For $(4,1,0)$:

2. For $(1,3,0)$:

3. For $(2,1,1)$:

4. For $(0,1,2)$:

Summing these individual coefficients gives the total coefficient of $x^{11}$:

The final coefficient is 1113.