Correct way to read a given permutation cipher? In some literature I am reading – where they are explaining the permutation cipher – they have two examples, which encrypt a small block of text. But they seem to contradict each other: 

First example: 


  permutation: ${1\text{ }2\text{ }3 \choose 3\text{ }1\text{ }2}$, plaintext: VEN, ciphertext =NVE


Second example:


  permutation: ${1\text{ }2\text{ }3 \choose 2\text{ }3\text{ }1}$, plaintext: TEN, ciphertext = NTE


Am I missing something? The second example makes sense to me, but not the first. Why isn’t the ciphertext for the first example not ENV?
The permutations in your question are given in Cauchy's two-line notation, where the upper line gives the input index to the permutation function, and the lower line gives the resulting permuted index.

For example, the definition $$\sigma = {1\ 2\ 3 \choose 3\ 1\ 2}$$ means the same as $$\sigma(1) = 3,\quad \sigma(2) = 1,\quad \sigma(3) = 2.$$  Thus, if we apply this permutation $\sigma$ to a tuple $x = (x_1, x_2, x_3)$, we obtain the permuted tuple $$x_\sigma  = \sigma \cdot x = (x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}) = (x_3, x_1, x_2).$$

In particular, if $x = (x_1, x_2, x_3) = (\text{V}, \text{E}, \text{N})$, then $x_\sigma = (x_3, x_1, x_2) = (\text{N}, \text{V}, \text{E})$.

This is easy visualize if you simply replace the indices $1$, $2$ and $3$ in the definition of the permutation with the elements $x_1 = \text{V}$, $x_2 = \text{E}$ and $x_3 = \text{N}$ to be permuted, giving $${x_1\ x_2\ x_3 \choose x_3\ x_1\ x_2} = {\text{V E N} \choose \text{N V E} }.$$



The second example in your question appears to be incorrect: given the permutation $$\sigma' = {1\ 2\ 3 \choose 2\ 3\ 1}$$ and the input $x' = (x'_1, x'_2, x'_3) = (\text{T}, \text{E}, \text{N})$, the output should be $$x'_{\sigma'}  = \sigma' \cdot x' = (x'_{\sigma'(1)}, x'_{\sigma'(2)}, x'_{\sigma'(3)}) = (x'_2, x'_3, x'_1) = (\text{E}, \text{N}, \text{T}),$$ not $(x'_3, x'_1, x'_2) = (\text{N}, \text{T}, \text{E})$.  It would, however, be correct if we replaced the permutation $\sigma'$ with its inverse $$\sigma'^{-1} = {2\ 3\ 1 \choose 1\ 2\ 3} = {1\ 2\ 3 \choose 3\ 1\ 2}$$ (which, as it happens, equals the first permutation $\sigma$) obtained by swapping the rows and (optionally) re-sorting the columns in the two-line definition of $\sigma'$.