Final answer:
The probability of digits 1 and 2 being neighbors in a random arrangement of numbers from 1 to n is 2/n, and the probability of 1, 2, and 3 being neighbors is 6/(n*(n-1)).
Step-by-step explanation:
To answer this mathematics question on probability, we first need to understand the arrangements of 1 and 2 as neighbors, and then of 1, 2, and 3 as neighbors in a random order.
(a) Considering numbers 1 through n are arranged randomly, if we want 1 and 2 to be neighbors, we can treat them as a single unit. There are (n-1)! ways to arrange the n-1 units (the combined 1 and 2, plus the n-2 other numbers). But, since 1 and 2 can be in two orders (1 then 2, or 2 then 1), we have 2 * (n-1)! arrangements. The total number of ways to arrange n numbers is n!, so the probability is P(1 and 2 together) = 2 * (n-1)! / n! = 2/n.
(b) For 1, 2, and 3 being next to each other, we treat them as a single unit again. Now we have (n-2)! ways to arrange them, and since they can be in 3! = 6 different orders, there are 6 * (n-2)! arrangements. Thus, the probability is P(1, 2, and 3 together) = 6 * (n-2)! / n! = 6/(n*(n-1)).