5. The numbers 1 through 12 are placed in a row. The number of ways this can be done is $12 \times 11 \times 10 \times \cdots \times 1$. We require that in such a row, there is exactly one number that is smaller than the number that directly precedes it.

How many of the $12 \times 11 \times 10 \times \cdots \times 1$ rows satisfy this requirement?
5. A sequence that meets the given condition is called 'good'. We will determine the number of good sequences indirectly by first counting something else.

We first look at the possible ways to color the numbers from 1 to 12, each with blue or red. Since there are two possibilities for each of the 12 numbers: either red or blue, there are $2^{12}$ colorings. We call such a coloring 'good' if there is at least one red and at least one blue number, and the largest red number is greater than the smallest blue number. There are exactly 13 colorings that are not good, namely when no number is red, or when the red numbers are exactly the numbers 1 to $k$ for some $k=1,2, \ldots, 12$ (if $k=12$, there are no blue numbers). In total, there are thus $2^{12}-13=4083$ good colorings.

We now show that the number of good sequences is equal to the number of good colorings by pairing them one-to-one. Take a good sequence and let $a$ be the number that is smaller than its predecessor in the sequence. We now color the numbers that come before $a$ red and color $a$ and the numbers that come after $a$ blue. This gives a good coloring.

For example, if we take the good sequence $1,3,4,5,8,9,2,6,7,10,11,12$, then $a$ is the number 2. We then color 1, 3, 4, 5, 8, and 9 red and color 2, 6, 7, 10, 11, and 12 blue. This is a good coloring because $9>2$.

Every good coloring is obtained this way exactly once. Take any good coloring, and we can find the corresponding good sequence as follows: write down all the red numbers in ascending order, followed by the blue numbers, also in ascending order.

In total, there are thus 4083 sequences that meet the condition.