Example 5. How many different permutations of letters can be made from the words: zamok, rotor, topor, kolokol?
Solution. In the word "замок" (lock), all letters are different, and there are five of them. According to formula (1.3.1), we get

$$
P_{5}=5!=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5=120
$$

In the word "ротор" (rotor), consisting of five letters, the letters $p$ and о are repeated twice. To count the different permutations, we use formula (1.3.7). For $n=5, n_{1}=2, n_{2}=2$, by this formula we find

$$
P_{5}(2 ; 2)=\frac{5!}{2!\cdot 2!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}{1 \cdot 2 \cdot 1 \cdot 2}=30
$$

In the word "топор" (hatchet), the letter о is repeated twice, so

$$
P_{5}(2)=\frac{5!}{2!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}{1 \cdot 2}=60
$$

In the word "колокол" (bell), consisting of seven letters, the letter $к$ appears twice, the letter $o$ - three times, and the letter $л$ - twice. According to formula (1.3.7) for $n=7, n_{1}=2, n_{2}=3, n_{3}=2$, we get

$$
P_{7}(2 ; 3 ; 2)=\frac{7!}{2!\cdot 3!\cdot 2!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{1 \cdot 2 \cdot 1 \cdot 2 \cdot 3 \cdot 1 \cdot 2}=210
$$

17р and mer 6. On five identical cards, the letters $U, K, M$, $H, C$ are written. The cards are shuffled and randomly laid out in a row. What is the probability that the word "МИНСК" (MINSK) will be formed?

Solution. From five different elements, $P_{5}$ permutations can be formed: $P_{5}=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5=120$. Therefore, there will be 120 equally possible outcomes, and only one of them is favorable to the given event. Thus,

$$
P=\frac{1}{120}
$$