How many Markov chains are there for 2 states in 1, 2 and 3 steps? wiki uses this example to illustrate Markov chains.

The probabilities of weather conditions (modeled as either rainy or
sunny), given the weather on the preceding day, can be represented by
a transition matrix:
${\displaystyle P={\begin{bmatrix}0.9&0.1\\0.5&0.5\end{bmatrix}}}$
The matrix P represents the weather model in which a sunny day is 90% likely to be followed by another sunny day, and a rainy day is 50% likely to be followed by another rainy day. The columns can be labelled "sunny" and "rainy", and the rows can be labelled in the same order.
The weather on day 1 is known to be sunny. This is represented by a vector in which the "sunny" entry is 100%, and the "rainy" entry is 0%:
${\displaystyle \mathbf {x} ^{(0)}={\begin{bmatrix}1&0\end{bmatrix}}}$

for day n + 1(Note: the original value on wiki is n, which seems to be incorrect)

${\mathbf  {x}}^{{(n)}}={\mathbf  {x}}^{{(0)}}P^{n}$

The superscript (n) is an index, and not an exponent.
In the particular case, the state space of the chain is {rainy , sunny}
how many Markov chains are there respectively on day1, day2 and day3?
for example, on day1
${\displaystyle \Pr(X_0=sunny) = 1,}$ ${\displaystyle \Pr(X_0=rainy) = 0,}$
how many Markov chains are there on day1, 1 or 2?
how many Markov chains are there on day2 and day3 respectively?
 A: Let $\{X_n:n=0,1,2\ldots\}$ be a Markov chain with transition matrix $P$. Then the $(i,j)$-entry of $P$ is the probability of transitioning from state $i$ to state $j$ in $n$ steps:
$$
P_{ij} = \mathbb P(X_n = j\mid X_0 = i).
$$
Here we have
$$
P = \left(
\begin{array}{cc}
 \frac{9}{10} & \frac{1}{10} \\
 \frac{1}{2} & \frac{1}{2} \\
\end{array}
\right),\quad P^2 = \left(
\begin{array}{cc}
 \frac{43}{50} & \frac{7}{50} \\
 \frac{7}{10} & \frac{3}{10} \\
\end{array}
\right),
$$
and in general
$$
P^n = \left(
\begin{array}{cc}
 \frac{1}{6} \left(\left(\frac{2}{5}\right)^n+5\right) & \frac{1}{6} \left(1-\left(\frac{2}{5}\right)^n\right) \\
 \frac{5}{6} \left(1-\left(\frac{2}{5}\right)^n\right) & \frac{1}{3} \left(\frac{2}{5}\right)^{n-1}+\frac{1}{6} \\
\end{array}
\right),
$$
so that
$$
\lim_{n\to\infty} P^n = \begin{pmatrix}
\frac56&\frac16\\\frac56&\frac16.
\end{pmatrix}
$$
Hence $X_n$ has limiting distribution
$$
\lim_{n\to\infty} \mathbb P(X_n = j) = \begin{cases}\frac56,&j=\mathrm{sunny}\\
\frac16,&j=\mathrm{rainy}
\end{cases},
$$
independent of the distribution of $X_0$.
