14. The knut ciphers in the library catalog consist of six digits and do not start with the digit 0. A reader is looking for the cipher of the book they need in the catalog. What is the probability that all the digits of the cipher will be different?
S o l u t i o n. Let us denote the event: $\boldsymbol{A}$ - all the digits of the code the reader is looking for are different.

We will find the probability of event $A$ by applying formula (1).

Let us determine the numbers $m$ and $n$ that are part of this formula.

There are ten elements in total - ten digits: $0,1, \ldots, 9$. In the book's code, digits occupy $1-\mathrm{e}, 2-\mathrm{e}, \ldots, 6$-th positions, which we will number I, II,..., VI.

There are ten ways to fill position I with the digits $0,1, \ldots, 9$ and ten such ways to fill the remaining positions II-VI. Each of the ten ways to fill position I can be combined with any of the ten ways to fill positions II-VI, so the total number of outcomes of the trial is $n=10^{6}$.

The following are possible ways to fill six positions with different digits: $2-5-6-9-8-7,2-6-5-9-8-7,0-4-6-7-5-3,1-4-6-7-5-3$ and others (some of which are combinations starting with the digit 0). The number of these combinations is $m_{1}$, each of which includes 6 elements from the available 10. The combinations differ from each other both in the elements themselves and in the order of the elements (the order of the elements is important). Therefore, the considered combinations are permutations of ten elements taken six at a time: $m_{1}=A_{10}^{6}$.

According to the condition, the code does not start with the digit 0. Therefore, from the number $m_{1}$ of combinations with different digits, we need to exclude the number $m_{2}$ of combinations that have different digits and start with zero. If we assume that position I is occupied by the digit 0, then the remaining five positions are filled with different digits from the remaining nine; therefore, $m_{2}=A_{9}^{5}$.

Thus,

$m=m_{1}-m_{2}=A_{10}^{6}-A_{9}^{5}=10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5-9 \cdot 8 \cdot 7 \cdot 6 \cdot 5=136080$.

The probability of event $A$ is $\quad P(A)=\frac{m}{n}=\frac{136080}{10^{6}}=0.13608$.