Find the binomial coefificient of $x^8$ in $(1+x^2-x^3)^9$

Question

I was trying to solve it using the multinomial theorem.

I was trying to find which combinations could give me such $x^8$

and I came to the conclusion that it only occours when i take $(x^2)^4$ or $(-x^3)^2*(x^2)^1$

therefore: $\binom{9!}{0!*4!*0!} + \binom{9!}{0!*1!*2!}$

It just seems just too big of a number.. what am I missing?

Answer 1

The required coefficient should be

$$\binom90\binom94+\binom92\binom71$$

Using binomial expansion of $$((1+x^2)+(-x^3))^9$$

Answer 2

Using multinomial theorem you should get after your correct calculations of the possible combinations for getting the exponent $8$ the following:

• Using the color code: $\color{green}{1} + \color{blue}{x^2} \color{orange}{-x^3}$

$$\binom{9}{\color{green}{5},\color{blue}{4},\color{orange}{0}} + \color{orange}{(-1)^2}\binom{9}{\color{green}{6},\color{blue}{1},\color{orange}{2}} = \frac{9!}{5!\cdot 4! \cdot 0!} + \frac{9!}{6!\cdot 1! \cdot 2!} = \boxed{378}$$

Answer 3

A comparison. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$.

Multinomial theorem:

\begin{align*} \color{blue}{[x^8]}&\color{blue}{\left(1+x^2-x^3\right)^9}\\ &=[x^8]\sum_{{k_1+k_2+k_3}\atop{k_1,k_2,k_3\geq 0}}\binom{9}{k_1,k_2,k_3}1^{k_1}\left(x^2\right)^{k_2}\left(-x^3\right)^{k_3}\\ &=[x^8]\sum_{{k_1+k_2+k_3}\atop{k_1,k_2,k_3\geq 0}}\binom{9}{k_1,k_2,k_3}(-1)^{k_3}x^{2k_2+3k_3}\\ &=\binom{9}{5,4,0}(-1)^0+\binom{9}{6,1,2}(-1)^2\tag{1}\\ &=\frac{9!}{5!4!0!}+\frac{9!}{6!1!2!}\\ &=126+252\\ &\,\,\color{blue}{=378} \end{align*}

Comment:

• In (1) we select the coefficient of $x^8$. We also note the sum of the entries of the lower indices of the multinomial coefficient is $9$ (i.e. $5+4+0=6+1+2=9$).

Binomial theorem (twice):

\begin{align*} \color{blue}{[x^8]}&\color{blue}{\left(1+x^2-x^3\right)^9}\\ &=[x^8]\sum_{k=0}^9\binom{9}{k}\left(x^2+x^3\right)^k\\ &=[x^8]\sum_{k=0}^9\binom{9}{k}x^{2k}\left(1+x\right)^k\\ &=\sum_{k=0}^4\binom{9}{k}x^{2k}[x^{8-2k}]\left(1+x\right)^k\tag{2}\\ &=\sum_{k=0}^4\binom{9}{k}\binom{k}{8-2k}\tag{3}\\ &=\binom{9}{3}\binom{3}{2}+\binom{9}{4}\binom{4}{0}\tag{4}\\ &=252+126\\ &\,\,\color{blue}{=378} \end{align*}

Comment:

• In (2) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and set the upper limit of the sum to $4$ since the exponent of $x^{8-2k}$ is non-negative.

• In (3) we select the coefficient of $x^{8-2k}$.

• In (4) we select the non-zero terms of (3) noting that $\binom{p}{q}=\frac{p(p-1)\cdots(p-q+1)}{q!}=0$ if $q>p$.