# Distinct terms in a binomial expansion

Find the total number of distinct terms in the expansion of $(2-2x+x^2)^9$.

According to the solution of the above problem , the following sum can be written as :

$$(2-2x+x^2)^9 = \left[ 1 + (1-x)^{\color{red}2}\right]^9 = \sum_{r=1}^9 \binom{9}{r}(1-x)^{2r}$$

Hence the highest degree will be $19$. So there are 18 distinct terms.

However we also know that the number of distinct terms in a multinominal expansion in given by $\binom{n+r-1}{r-1}$ . However we cannot use this formula here, because the actual number of terms will be lesser than that given by this formula. Could you please explain why does this happen and in what cases can we use this formula ? Thank you for your help.

• When you talk about the number of "terms" in the expansion, you should be clear whether you mean before or after simplifications and what counts as a "term" in the first place. One could argue that $(1+1)^2=1+1+1+1$ has four terms by using the "foil" method. One could also argue that $(1+1)^2=1+2+1$ by binomial expansion and has three terms. One could also argue that $(1+1)^2=4$ has only a single term. – JMoravitz Jan 19 '18 at 21:02
• @JMoravitz so could you please tell you kinds of terms in the above expansion would get clubbed thus reducing the number of terms? – Aditi Jan 19 '18 at 21:06
• In your case, if you were to expand $(\color{blue}{2}\color{green}{-2x}+\color{purple}{x^2})^9$ via multinomial expansion, you will have for example a term involving $\color{blue}{2}^8\cdot (\color{purple}{x^2})^1$. You will also have a term involving $\color{blue}{2}^7\cdot (\color{green}{-2x})^2$. These both will wind up as multiples of $x^2$ in the end and so can combine, the question is whether you consider these separate or not. They certainly would have been different in the expansion of $(a+b+c)^9$, but some would argue not in the expansion of $(1+1+1)^9$ nor in $(2-2x+x^2)^9$ – JMoravitz Jan 19 '18 at 21:07
• @JMoravitz thank you very much ! – Aditi Jan 20 '18 at 3:38

Highest degree will be $9 \cdot 2 = 18$ and total number of terms $19$.
In other words, if $p(x)$ is a polynomial of degree $d$, and $n \in \mathbb{N}$ then the degree of $p(x)^n$ is $d \cdot n$.
• @stevengregory yes, but not the highest-degree term, which can be $0$ if and only if $d=0$ or $n=0$. – gt6989b Jan 21 '18 at 2:11
• But he asked for the "total number of distinct terms". So 19 is only an upper bound until you can show that every $a_ix^i$ has $a_i \ne 0$ – steven gregory Jan 21 '18 at 3:30