85. What is the probability that $C_{n}^{7}$, where $n$ is a randomly chosen integer greater than seven, is divisible by 7? What is the probability that $C_{n}^{7}$ is divisible by 12?
85. $C_{n}^{7}=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}$; therefore, the probability that $C_{n}^{7}$ is divisible by 7 is equal to the probability that the product $n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ is divisible by 49. But the latter, obviously, will occur if and only if $n$ gives one of the remainders $0,1,2,3,4,5$ or 6 when divided by 49. Out of any 49 consecutive integers, exactly 7 satisfy this condition. Hence, the desired probability is $\frac{7}{49}=\frac{1}{7}$.

The probability that $C_{n}^{7}$ is divisible by 12 is obviously the same as the probability that the product in the numerator $n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ is divisible by $64 \cdot 27$. Out of any seven consecutive numbers, 2 are necessarily divisible by 3; if one of them is divisible by 9, then the entire product is necessarily divisible by 27. Therefore, if $n$ has the form $9k + r$, where $r=0,1,2,3,4,5$ or 6, then the product $n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ is divisible by 27. (The case where three factors of our product are divisible by 3 is also included here, since among three consecutive numbers divisible by three, one is necessarily divisible by 9.)

Further, out of our seven consecutive integers, $n-1, n-3$ and $n-5$ or $n, n-2, n-4$ and $n-6$ are divisible by 2. In the first case, out of the three consecutive even numbers $n-1$, $n-3$ and $n-5$, $n-3$ or $n-1$ and $n-5$ are divisible by 4. If $n-1$ and $n-5$ are divisible by 4, then out of these two consecutive numbers divisible by four, at least one is divisible by 8; therefore, if $n-1$ is divisible by 4, i.e., $n$ has the form $4l + 1$, then our product is divisible by $2 \cdot 4 \cdot 8 = 64$. If $n-3$ is divisible by 4, then for the product to be divisible by 64, $n-3$ must be divisible by 16, i.e., $n$ must have the form $16l + 3$. Finally, if $n, n-2, n-4$ and $n-6$ are divisible by 2 (if $n = 2l$ is even), then out of these four consecutive even numbers, two must be divisible by 4 and, therefore, in this case, our product is necessarily divisible by $2 \cdot 2 \cdot 4 \cdot 4 = 64$ (it is easy to see that in this case it must be divisible even by 128).

Thus, $n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ is divisible by 64 when $n$ has the form $4l + 1$ or $16l + 3$, or $2l$, and only in these cases, in other words, when $n$ has the form $16l + s$, where $s=0,1,2,3,4,5,6,8,9,10$, 12, 13, or 14.

Thus, $C_{n}^{7}$ is divisible by 12 if $n$ has the form $9k + r$, where $r$ has one of the above seven values and simultaneously has the form $16l + s$, where $s$ has one of the above 13 values. Accordingly, all $n$ such that $C_{n}^{7}$ is divisible by 12 must give one of $7 \cdot 13 = 91$ remainders when divided by $9 \cdot 16 = 144$ (for example, if $n$ gives a remainder of 8 when divided by 16 and a remainder of 5 when divided by 9, then $n$ must have the form $144m + 104$: among the numbers of the form $144m + t$, where $t=8,24,40,56,72,88,104,120,136$, only the numbers of the form $144m + 104$ give a remainder of 5 when divided by 9).

Hence, the desired probability is $\frac{91}{144} \approx 0.63$.