Mr. Chrt had five dogs in his dog team - Alík, Brok, Muk, Raf, and Punt. He was thinking about how he could harness the dogs in a row behind him so that Alík would be in front of Punt.

In how many ways could Mr. Chrt do this?

(L. Hozová)

Hint. On which positions could Alík and Punt be harnessed?
If Alík was first, Punta could be second, third, fourth, or fifth:

$$
A P * * * \quad A * P * * \quad A * * P * \quad A * * * P
$$

If Alík was second, Punta could be third, fourth, or fifth:

$$
* A P * * \quad * A * P * \quad * A * * P
$$

If Alík was third, Punta could be fourth or fifth:

$$
* * A P * \quad * * A * P
$$

If Alík was fourth, Punta had to be fifth:

$$
* * * A P
$$

Thus, Mr. Chrt had 10 options for harnessing Alík and Punta in the required manner.

In each of these cases, the remaining three dogs could be harnessed in the unoccupied positions (marked with *) in any order. This can be done in 6 ways: one dog can go to any of the three free positions, the second dog to any of the two remaining positions, and the third dog has no choice and must go to the last free position (thus $6=3 \cdot 2$).

Mr. Chrt could harness his dogs in a total of 60 ways (since $10 \cdot 6=60$).

Note. The previous calculation of all possible orders of three elements can be generalized for any number; the resulting number is called a factorial and is denoted by an exclamation mark (here $3 !=3 \cdot 2 \cdot 1=6$).

All five of Mr. Chrt's dogs can be harnessed in a total of $5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ ways. In half of the cases, Alík stands in front of Punta, and in the other half, it is the opposite (it is enough to swap these two dogs and leave the others in their places). Thus, Alík can be in front of Punta in 60 cases.