27.7. (New York, 76). The sequence ( $i_{1} ; i_{2} ; \ldots ; i_{n}$ ) is formed from the first $n>3$ natural numbers, arranged in a random order. What is the probability that for all $k=1,2, \ldots, n$ the inequality $i_{k} \geqslant k-3$ holds?
27.7. Let's find the number of sequences ( $i_{1} ; \ldots ; i_{n}$ ), for which $i_{k} \geqslant k-3$ for $k=1,2, \ldots, n$. The number $i_{n}$ can take four values: $n, n-1, n-2, n-3$. The number $i_{n-1}$ can take five values: $f, n-1, n-2, n-3, n-4$, except for the value already occupied by the number $i_{n}$. Thus, the number $i_{n-1}$ can also take 4 values. Similarly, each of the numbers $i_{n-2}, \ldots, i_{4}$ can take 4 values. The numbers $i_{1}, i_{2}, i_{3}$ can be chosen arbitrarily from the three values remaining after selecting the numbers $i_{n}, \ldots, i_{4}$. Thus, among all $n$! possible sequences, there are exactly $4^{n-3} \cdot 3$! sequences that satisfy the required condition. Therefore, the desired probability is " 4 " $n$ - $3.3!/ n!$.