Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{10}$ in the expansion of $(1+x{)}^2(1+x^2{)}^3(1+x^3{)}^4$ is equal to :-

Imagine you are standing before a massive, intimidating wall of algebra: $(1+x{)}^2(1+x^2{)}^3(1+x^3{)}^4$. The question asks for the coefficient of $x^{10}$.

In the world of JEE Advanced, brute force is rarely the path to victory. The path to victory is elegance. We will navigate this maze by focusing only on the terms that contribute to our target power.

Think of this expression as three distinct baskets. To form the final expansion, you must reach into each basket, pull out exactly one term, and multiply them together.

The first basket, $(1+x{)}^2$, expands to $1+2x+x^2$. The powers available here are $\{0,1,2\}$.

The second basket, $(1+x^2{)}^3$, expands to $1+3x^2+3x^4+x^6$. The powers available here are $\{0,2,4,6\}$.

The third basket, $(1+x^3{)}^4$, expands to $1+4x^3+6x^6+4x^9+x^{12}$. The powers available here are $\{0,3,6,9\}$ (we ignore $12$ as it exceeds our target).

We need the sum of the exponents to be $10$. Let $a$ be the exponent chosen from the first basket, $b$ from the second, and $c$ from the third.

Our mission is to solve the equation:

Subject to the constraints:

We will fix $c$ from the third basket because it has the largest steps, which limits our search space significantly.

Case 1: Let $c=3$. The equation becomes $a+b=7$. Given $b\in \{0,2,4,6\}$, the only solution is $b=6$ and $a=1$. The coefficient is $(2)\times (1)\times (4)=8$.

Case 2: Let $c=6$. The equation becomes $a+b=4$. If $b=4$, then $a=0$. The coefficient is $(1)\times (3)\times (6)=18$. If $b=2$, then $a=2$. The coefficient is $(1)\times (3)\times (6)=18$.

Case 3: Let $c=9$. The equation becomes $a+b=1$. Since $b$ must be even, $b$ must be $0$, which forces $a=1$. The coefficient is $(2)\times (1)\times (4)=8$.

We have exhausted all possibilities to construct $x^{10}$. We now sum the coefficients:

We didn't expand a single complex polynomial. We used logic, constraints, and systematic counting to arrive at the final answer of 52.