Arrangement of the words "SWEETNESS" in a $3 × 3$ matrix Number of ways in which all the letters of the words "SWEETNESS" can be arranged in matrix of $3 × 3$ such that no letter in any row or column is repeated:
My approach is as follow

The image is one of the possible arrangement where rows and column are different, how do we find the arrangement.  In the figure $a_{ij}$ represent the matrix in rows and column arrangement
 A: We can find the number of ways to arrange the letters by putting letters in in a sensible order and analyzing the number of choices we have at each step. Starting with the blank matrix:
$$\left[\begin{matrix}
\square&\square&\square\\
\square&\square&\square\\
\square&\square&\square
\end{matrix}\right]$$
There are clearly $9$ ways to place the first $S$ into the matrix. I'll be putting letters in the example matrix arbitrarily, but each step should be valid for any previous choices.
$$\left[\begin{matrix}
S&{\color{red}\square}&{\color{red}\square}\\
{\color{red}\square}&{\color{}\square}&\square\\
{\color{red}\square}&\square&\square
\end{matrix}\right]$$
Note that after blocking out the row and column of our first $S$, there are $4$ choices for the next.
$$\left[\begin{matrix}
S&{\color{red}\square}&{\color{red}\square}\\
{\color{red}\square}&{\color{red}\square}&S\\
{\color{red}\square}&\square&{\color{red}\square}
\end{matrix}\right]$$
Now there is only one square which is valid for the last $S$.
$$\left[\begin{matrix}
S&{\color{}\square}&{\color{}\square}\\
{\color{}\square}&{\color{}\square}&S\\
{\color{}\square}&S&{\color{}\square}
\end{matrix}\right]$$
Now note that the $S$'s cannot be distinguished from each other. Thus each possible configuration has been counted once for each order of the $S$'s, so we must divide by $3!=6$.
Next, we have $6$ options for the first $E$.
$$\left[\begin{matrix}
S&E&{\color{red}\square}\\
{\color{}\square}&{\color{red}\square}&S\\
{\color{}\square}&S&{\color{}\square}
\end{matrix}\right]$$
Note that since there is an $S$ in every row and column, there is one $S$ each in the same row and column as the $E$. Thus there is $1$ $S$ in the $4$ remaining squares. But as you can see below, one of the configurations leaves no options for the last $E$. So we have $2$ choices here.
$$\left[\begin{matrix}
S&E&{\color{red}\square}\\
{\color{orange}\square}&{\color{red}\square}&S\\
{\color{orange}E}&S&{\color{orange}\square}
\end{matrix}\right]\qquad\left[\begin{matrix}
S&E&{\color{red}\square}\\
{\color{lime}\square}&{\color{red}\square}&S\\
{\color{orange}\square}&S&{\color{lime}E}
\end{matrix}\right]$$
Again we have only one choice for the last $E$, and must account for $3!=6$ double-countings.
$$\left[\begin{matrix}
S&E&{\color{}\square}\\
E&{\color{}\square}&S\\
{\color{}\square}&S&E
\end{matrix}\right]$$
Finally, we have a $N$, $T$, and $W$ to place in the remaining $3$ places. There are $3!$ ways to do this:
$$\left[\begin{matrix}
S&E&{\color{lime}T}\\
E&{\color{lime}W}&S\\
{\color{lime}N}&S&E
\end{matrix}\right]$$
Going back through all our choices, we get that the total number of valid matricies is:
$$N=\frac{9\cdot4\cdot1\cdot6\cdot2\cdot1\cdot3!}{3!\cdot3!}=72$$
A: There are 3 S and 3 E, so first put one S and one E in each group such there is not duplicates in the columns:
$$[S, E, \cdot][S,\cdot ,E ][\cdot, S,E ]$$
Then you need to situate only the three letters left WTN. There are $3!$ ways to situate those three letters, by wich there are $3!$ ways to arrange the three rows, by wich there are $2!$ ways to arrange letters E and S at the same time, so the number you are looking for is $$2\cdot3!^2=72$$
A: First choose the 3 positions of "S" by choosing a permutation$~\pi$ of$~\{1,2,3\}$ (using it for instance to map row indices to the corresponding column with an "S"), which can be done in $3!=6$ ways. Then similarly select the $3$ positions with "E" by choosing another such permutation$~\sigma$; here the choice is limited by the fact that $\sigma(i)\neq\pi(i)$ for $i=1,2,3$, which means that $\pi^{-1}\sigma$ has no fixed points, and there are $2$ possibilities (the cyclic permutations of order$~3$) for $\pi^{-1}\sigma$, and therefore as many for $\sigma$ (one could use the formula for derangement counting here but just listing is easier). Finally fill in the remaining $3$ letters in $3!$ ways, for $72$ solutions in all.
