Mathematics - Binomial Theorem Question with Solution | TestHub

We need to find the coefficient of $x^7$ in ${\left(1-x+2x^3\right)}^{10}$.

We know that for ${\left(x+y+z\right)}^n$, 

$T_n=\frac{n!}{a!b!c!}{\left(x\right)}^a{\left(y\right)}^b{\left(z\right)}^c$ such that $a+b+c=n$

Now for ${\left(1-x+2x^3\right)}^{10}$

$T_n=\frac{10!}{a!b!c!}{\left(1\right)}^a{\left(-x\right)}^b{\left(2x^3\right)}^c$

$=\frac{10!}{a!b!c!}{\left(-1\right)}^b{2}^c{x}^{b+3c}$. with $a+b+c=10$

Here we need coefficient of $x^7$.

Hence the combinations would be

$\begin{array}{ccc}a& b& c\\ 3& 7& 0\\ 5& 4& 1\\ 7& 1& 2\end{array}$ 

which satisfies both $a+b+c=10$ and $b+3c=7$.

Hence, coefficient of $x^7$ will be,

 $=\frac{10!}{2!1!7!}{\left(2\right)}^2{\left(-1\right)}^1$

$+\frac{10!}{1!4!5!}{\left(2\right)}^1{\left(-1\right)}^4+\frac{10!}{0!7!3!}{\left(2\right)}^0{\left(-1\right)}^7$

$=960$