, 2020

(a) Each circle can be colored with two different colors independently of the coloring of the other circles. Therefore, according to the multiplicative principle, the different colorings are
$$\underbrace{2 \cdot 2 \cdot 2 \dotsm 2}_{99} = 2^{99}.$$

(b) We consider the coloring $X: 1\ 2\ 3\ \dots\ 50\ \underbrace{51\ \dotsm\ 98\ 99}_{x}$, where $x$ is the number of red circles among $1$ to $50$, and $y$ is the number of red circles among $51$ to $99$. The coloring $X$ is «good» if $x > y$, whereas the coloring $X$ is *no good* if $x \le y$.

Let $A$ be the set of «good» colorings and $B$ the set of “no good” colorings. We will prove that each element of the set $A$ corresponds to an element of $B$ and vice versa.

Indeed, if $X: 1\ 2\ 3\ \dots\ 50\ \underbrace{51\ \dotsm\ 98\ 99}_{x}$ is in $A$, then $x > y$. By changing the color of each circle, we find the coloring
$$Y: \underbrace{1\ 2\ 3\ \dots\ 50}_{50-x}\ \underbrace{51\ \dotsm\ 98\ 99}_{49-y},$$
which belongs to $B$, because
$$x > y \Rightarrow -x < -y \Rightarrow 49 - x < 49 - y \Rightarrow 50 - x \le 49 - y.$$

Conversely, if $X: 1\ 2\ 3\ \dots\ 50\ \underbrace{51\ \dotsm\ 98\ 99}_{x}$ is in $B$, then $x \le y$. By changing the color of each circle, we find the coloring
$$Y: \underbrace{1\ 2\ 3\ \dots\ 50}_{50-x}\ \underbrace{51\ \dotsm\ 98\ 99}_{49-y},$$
which belongs to the set $A$ because
$$x \le y \Rightarrow -x \ge -y \Rightarrow 49 - x \ge 49 - y \Rightarrow 50 - x > 49 - y.$$

Therefore, between the sets $A$ and $B$ there exists a $1$-$1$ correspondence and so the two sets have the same number of elements, that is, each of them has
$$\frac{2^{99}}{2} = 2^{98} \text{ elements.}$$