All plus and equal is in the sense of $Z_{10}$ or $Z_{10}^4$

Given $M_0 = \begin{pmatrix}0&0&0&0\end{pmatrix}$ and $M_1$ and $C =\begin{pmatrix}0&0&0&0\end{pmatrix}$ :

$\begin{aligned}C_0 = \begin{pmatrix}\pi(d)&\pi(d)&\pi(d)&\pi(d)\end{pmatrix}\\C_1 = \begin{pmatrix}\pi(d)&\pi(d - 1)&\pi(d - 3)&\pi(d - 6)\end{pmatrix}\\Left = Pr(C_0 = C) = \sum_{d,\pi} Pr(d,\pi) Pr(C_0 = C \vert d ,\pi)\\Pr(C_0 = C \vert d,\pi) =\begin{cases}1&\pi(d) = 0\\0&else\end{cases}\\Right = Pr(C_1 = C) = \sum_{d,\pi} Pr(d,\pi) Pr(C_1 = C \vert d,\pi)\\Pr(C_1 = C \vert d,\pi) = \begin{cases}1&\pi(d) = \pi(d - 1)=\pi(d - 3) =\pi(d - 6) = 0\\0&else\end{cases}\end{aligned}$

The size of Key set of $(d, \pi)$ is $10 \cdot 10!$

As uniform selection , the Left is . Right is 0 because it is impossible $Pr(d, \pi) = \frac{1}{10 ∙ 10 !\frac{10 ∙ 9 !}{10 ∙ 10 !} = \frac{1}{\pi(d) =\pi(d - 1) =\pi(d - 3) =\pi(d - 6) = 0$

So $Left \ne Right$

Let the mapping from a key to the cipher string with a given $M$,

$Enc : K \rightarrow Enc(K) = \{ Enc(k) \vert k \in K \}$

Then the probability of a specific cipher text $C$ is:

$\begin{aligned}Pr(Enc(k) = C \vert k) = \begin{cases}1&Enc(k) = C\\0&else\end{cases}\\Pr(Enc(k) = C) = \sum_{k \in K} Pr(k) Pr(Enc(k) = C \vert k)\end{aligned}$

If distribution of $k$ is uniform, then

$Pr(Enc(k) = C) = \sum_{k \in K} Pr(k) Pr(Enc(k) = C \vert k) = \frac{1}{Count(K)} \sum_{k \in K} Pr(Enc(k) = C \vert k)$

The number $\sum_{k \in K} Pr(Enc(k) = C \vert k)$ may vary dependent on $M$; when $C$ is not in $Enc(K)$, this number is zero; when $C$ is in $Enc(K)$, the number typically vary with different $M$. If this number happens to be the same for any $C \in Enc(K)$ aka being secrecy, then it follows the distribution of the cipher text is also uniform:

$\frac{1}{CountK} \sum_{k \in K} Pr(Enc(k) = C \vert k) = \frac{Count(Enc(K))}{Count(K)}$

For example of Task 2:

$\begin{aligned}i = 1 \cdots n\\Count(K) = (2 n)!\\C_{\pi^{-1}(i)} = M_i\\C_{\pi^{-1}(i + n)} = M_{i + n}'= 1 - M_i\\Pr(Enc(k) = C \vert k) = \begin{cases}1&i = 1 \cdots n, C_{\pi^{-1}(i)} = M_i, C_{\pi^{-1}(i + n)} = 1 - M_i\\0&else\end{cases} \end{aligned}$

To evaluate $\sum_{k \in K} Pr(Enc(k) = C \vert k)$, consider all the index of 1 in $C$ being $t_1,t_2,\cdots,t_n$ and all the index of 0 in $C$ being $f_1,f_2,\cdots,f_n$ and all the index of 1 in $M'$ being $1_1,1_2,\cdots,1_n$ and all the index of 0 in $M'$ being $0_1,0_2,\cdots,0_n$

An example of n=4:

$\begin{aligned}C=\begin{pmatrix}1\\0\\1\\0\\0\\0\\1\\1\end{pmatrix},M'=\begin{pmatrix}1\\0\\0\\1\\0\\1\\1\\0\end{pmatrix},\begin{pmatrix}t_1&t_2&t_3&t_4\end{pmatrix} = \begin{pmatrix}1&3&7&8\end{pmatrix},\begin{pmatrix}f_1&f_2&f_3&f_4\end{pmatrix} =\begin{pmatrix}2&4&5&6\end{pmatrix}\\\begin{pmatrix}1_1&1_2&1_3&1_4\end{pmatrix} =\begin{pmatrix}1&4&6&7\end{pmatrix},\begin{pmatrix}0_1&0_2&0_3&0_4\end{pmatrix} =\begin{pmatrix}2&3&5&8\end{pmatrix}\end{aligned}$

Therefore, the count of $\pi$ satisfy the constraint of the message $M$ is $n!n!$ :

$\sum_{k \in K} Pr(Enc(k) = C \vert k) = n ! \cdot n !$

And in fact, the count number is irrelevant for any $M$:

For any $C$ in the range:

$Pr(Enc(k) = C) = \frac{n ! \cdot n !}{(2 n)!}$

Note that the size of the range is $C_n^{2 n}$ . Being uniform, it means each cipher text is probability of $\frac{1}{C_n^{2 n}} = \frac{n ! \cdot n !}{(2 n)!}$

For any $C$ not in the range: $Pr(Enc(k) = C) = 0$