Animated Solution for Mathematics - Binomial Theorem: If the coefficient of $a^7{b}^8$ in the expansion of $(a+2b+4ab{)}^{10}$ is $K\cdot 2^{16}$, then $K$ is equal to .

Welcome, fellow traveler on the JEE journey! Today, we are going to demystify a problem that often intimidates students at first glance: finding the coefficient of $a^7{b}^8$ in the expansion of $(a+2b+4ab{)}^{10}$.

It looks like a chaotic mess of variables, but I promise you, beneath the surface lies a beautiful, structured symmetry. Let us peel back the layers together.

The General Term

When we face an expression with three terms raised to a power, we step beyond the familiar Binomial Theorem and into the realm of the Multinomial Theorem. Imagine you have ten slots to fill, and you are distributing these slots among three different types of items: $a$, $2b$, and $4ab$.

Let $\alpha$ be the number of times we pick $a$, $\beta$ be the number of times we pick $2b$, and $\gamma$ be the number of times we pick $4ab$. The total number of picks must be ten, so our first constraint is $\alpha +\beta +\gamma =10$.

The general term is given by the formula:

This formula is the heartbeat of the problem.

The Algebraic Trap

Here is where many students stumble. They see $a$ and $b$ and try to equate $\alpha$ to $7$ and $\beta$ to $8$. But look closely at the term $(4ab{)}^{\gamma }$. It contains both $a$ and $b$!

When we expand the general term, we get:

We must group the like variables. The total power of $a$ is $\alpha +\gamma$, and the total power of $b$ is $\beta +\gamma$. The constant part is $2^{\beta }\cdot 4^{\gamma }$.

Since $4=2^2$, we can write this as $2^{\beta }\cdot 2^{2\gamma }=2^{\beta +2\gamma }$. Now, our expression is clean and ready for the next step.

The System of Equations

We want the coefficient of $a^7{b}^8$. This gives us a system of three linear equations:

1) $\alpha +\gamma =7$

2) $\beta +\gamma =8$

3) $\alpha +\beta +\gamma =10$

Solving this is pure joy. If we add (1) and (2), we get $\alpha +\beta +2\gamma =15$.

Now, subtract (3) from this sum: $(\alpha +\beta +2\gamma )-(\alpha +\beta +\gamma )=15-10$. The variables $\alpha$ and $\beta$ vanish, leaving us with $\gamma =5$.

With $\gamma =5$, we immediately find $\alpha =7-5=2$ and $\beta =8-5=3$. Our triplet is $(\alpha ,\beta ,\gamma )=(2,3,5)$.

The Final Calculation

Now, we substitute these values back into our coefficient expression:

We know $4^5=(2^2{)}^5=2^{10}$. So the constant part is $2^3\cdot 2^{10}=2^{13}$.

The factorial part is:

Our coefficient is $2520\cdot 2^{13}$. The problem asks for the form $K\cdot 2^{16}$. We need to extract $2^3$ (which is $8$) from $2520$.

$2520/8=315$. Thus, the coefficient is $315\cdot 2^3\cdot 2^{13}=315\cdot 2^{16}$.

Comparing this to $K\cdot 2^{16}$, we find $K=315$. You have conquered the multinomial expansion!