【Analysis and Solution】
(Method One)
$4 \div 3=1 \cdots \cdots 1, \quad 1+1=2$;
There must be one row with 2 coins;
There are 3 choices for the row with 2 coins;
Without loss of generality, consider the row with 2 coins to be the top row;
The 2 coins in the same row have $C_{3}^{2}=\frac{3 \times 2}{2 \times 1}=3$ choices;
Without loss of generality, consider the first row, with the first and second columns each having one coin;
Consider the remaining 3 coins, there are 5 choices through enumeration;
There are a total of $3 \times 3 \times 5=45$ different methods.
(Method Two)
Select any 4 cells from a $3 \times 3$ grid to place the coins, there are $C_{9}^{4}=126$ ways;
Among these, remove the ways of selecting any 4 cells from a $2 \times 3$ grid to place the coins, there are $C_{6}^{4} \times(3+3)=15 \times 6=90$ ways;
However, the ways of selecting any 4 cells from a $2 \times 2$ grid to place the coins have been removed twice, there are $C_{4}^{4} \times(3 \times 3)=1 \times 9=9$ ways; according to the principle of inclusion and exclusion, there are $126-90+9=45$ different methods.
【Question 14】
In the $3 \times 3$ table shown in the figure, 4 identical pieces are to be placed in the cells, with at most one piece per cell. If it is required that each row and each column must contain a piece, then there are $\qquad$ different ways to do this.