You are given the $4 \times 4$ grid below. Find the number of ways of placing $4$ counters in the squares (at most one counter per square), so that each row contains exactly one counter, and each column contains exactly one counter.
The first counter can be placed in any of the $4$ squares of the first row. Suppose it is placed in the square in the first column. Then, none of the remaining counters can be placed in the first column, so the second counter must be placed in the square in the second column, the third counter must be placed in the square in the third column, and the fourth counter must be placed in the square in the fourth column. The arrangement is then determined, so there is $1$ possible way.
If the first counter is placed in the square in the second column, then the second counter must be placed in the square in the first column. The third counter can be placed in the square in the third column or the square in the fourth column, and the fourth counter can be placed in the remaining square. This gives us $2$ possible ways.
If the first counter is placed in the square in the third column, then the second counter must be placed in the square in the fourth column. The third counter can be placed in the square in the first column or the square in the second column, and the fourth counter can be placed in the remaining square. This gives us $2$ possible ways as well.
Finally, if the first counter is placed in the square in the fourth column, then the second counter must be placed in the square in the third column. The third counter can be placed in the square in the first column or the square in the second column, and the fourth counter can be placed in the remaining square. This gives us $2$ possible ways.
In total, there are $1+2+2+2 = \boxed{7}$ possible ways.