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.
Highest degree will be $9 \cdot 2 = 18$ and total number of terms $19$.
You don't need your logic for that, just note that the highest degree term comes from taking highest-degree inside term (quadratic) to the power of the outer bracket (9).
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 $d \cdot n$.