Solution. Let's number the cards with letters:

| 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{a}$ | $\mathrm{a}$ | $\mathrm{T}$ | $\mathrm{H}$ | $\mathrm{r}$ | $\mathrm{r}$ |

The word "talant" will not change if the letters $a$ are swapped, 513246 but the arrangement of the cards will result in a different combination: $\max _{5}$ л $_{3}$ а $_{4}$ н. If the same is done with the letter $m$ in each of these two combinations, we will get 2 more different card combinations with the word "talant". Therefore, the word "talant" is favorable in 4 elementary outcomes. The total number of equally possible elementary outcomes is equal to the number of permutations of 6 elements: $n=6!=720$. Therefore, the desired probability is

$$
P=\frac{m}{n}=\frac{4}{720}=\frac{1}{180}
$$

Remark. This probability can also be found using formula (1.3.7), which for $n=6, n_{1}=1, n_{2}=1, n_{3}=2, n_{4}=2$ takes the form: $P_{6}(1,1,2,2)=\frac{6!}{1!\cdot 1!\cdot 2!\cdot 2!}=180$. Thus, $P=1 / 180$.
Example 8. On six cards of the same shape and size, the letters of the word talent are written - one letter on each card. The cards are thoroughly shuffled. They are randomly drawn and placed on the table one after another. What is the probability of getting the word $m a$ lant again?