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

Unlocking the Expansion of $(1+x+x^2{)}^{10}$

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a mathematical structure that often intimidates students.

When you see $(1+x+x^2{)}^{10}$, your first instinct might be to look for a binomial shortcut or to try and expand it manually. But take a breath. In the JEE Advanced arena, we don't brute-force; we use the elegance of the Multinomial Theorem.

This problem is a beautiful exercise in systematic thinking.

The Master Key

Imagine you are standing before a massive gate, and the Multinomial Theorem is your key. The general term for any expansion of the form $(a+b+c{)}^n$ is given by the formula:

Here, $n=10$, $a=1$, $b=x$, and $c=x^2$. When we substitute these into our formula, we get:

Notice how the $1^p$ simply vanishes into unity? That is the beauty of choosing your terms wisely. We are left with:

This is our target. We need the coefficient of $x^4$, which means we need the exponent $q+2r$ to equal $4$.

The Detective Work

Now, we enter the phase of the detective. We have two fundamental constraints that must be satisfied simultaneously:

1. The sum of the powers must equal the total power: $p+q+r=10$. 2. The power of $x$ must match our target: $q+2r=4$.

We also have the non-negotiable rule that $p,q,$ and $r$ must be non-negative integers. Do not rush; be systematic.

We will iterate through possible values of $r$. Why $r$? Because it has the largest coefficient in our constraint equation $q+2r=4$, making it the most restrictive variable.

If $r=0$: Then $q=4$. Substituting into the first constraint, $p+4+0=10$, which gives $p=6$. Our first triplet is $(6,4,0)$. If $r=1$: Then $q+2=4$, so $q=2$. Substituting into the first constraint, $p+2+1=10$, which gives $p=7$. Our second triplet is $(7,2,1)$. If $r=2$*: Then $q+4=4$, so $q=0$. Substituting into the first constraint, $p+0+2=10$, which gives $p=8$. Our third triplet is $(8,0,2)$.

What if $r=3$? Then $q+6=4$, implying $q=-2$. Since we cannot have a negative power, we stop. We have found all three possible scenarios.

The Victory Lap

Now, we calculate the coefficients for each case and sum them up. This is the victory lap.

For the first case $(6,4,0)$:

For the second case $(7,2,1)$:

For the third case $(8,0,2)$:

Finally, we add them together:

You see? By breaking the problem into manageable, logical steps, the complexity dissolves. You didn't just find the number 615; you navigated the logic of multinomial expansion. Keep this systematic approach in your toolkit, and no JEE problem will ever be too daunting.