6. From a deck of cards, 4 aces and 4 kings were removed. These cards were shuffled and laid out in a row. What is the probability that all 4 kings will be located next to each other?
S o l u t i o n. Let's denote the event: $A$ - 4 kings will be located next to each other. The probability of event $A$ will be found using formula (1).

The total number of $n$ possible outcomes of the experiment will be obtained similarly to how it is done in problem 5. There are 8 elements in total - 8 cards. These elements are represented in Fig. 3, marked with numbers from one to eight. Elements 1-4 are kings, represented by the symbols $\otimes$, and elements 5-8 are aces, represented by the symbols O. In forming different combinations, all 8 elements participate, and the combinations differ from each other only in the order of the elements; therefore, these combinations represent permutations of 8 elements. Using formula (2), we get: $n=P_{8}=8$ !.

| $\otimes$ | $\otimes$ | $\otimes$ | $\otimes$ | $\vee$ | $\mathrm{O}$ | $\vee$ | $\mathrm{O}$ | $\vee$ | $\mathrm{O}$ | $\vee$ | $\mathrm{O}$ | $\mathrm{v}$

Fig. 3

Let's find the number $m$ of outcomes favorable to event $A$.

Suppose the kings (elements 1-4) are placed in front of the aces, i.e., they occupy places 1-4. In this case, the aces (elements 5-8) can be permuted in all possible ways, occupying the remaining 4 free places. The number of these ways is $P_{4}=4$ !.

Consider what positions the elements 1-4 can occupy if they are next to each other and arranged in a certain order. On Fig. 3, the possible positions of these elements are shown by the signs $V$. As can be seen from the figure, there are 5 such positions. Therefore, the number of ways to arrange the four kings in a row so that they are next to each other and in a certain order is $4!.5=5!$.

The number of ways to permute the four kings among themselves is $P_{4}=4$ !.

Combining each first way of arranging the cards (kings are next to each other and in a certain order) with the second (kings are next to each other but in any order), we get: $m=5!4!$.

The sought probability of event $A$ is

$$
P(A)=\frac{m}{n}=\frac{5!\cdot 4!}{8!}=\frac{1}{14}
$$