Problem:
Let be the number of five-element subsets that can be chosen from the set of the first natural numbers so that at least two of the five numbers are consecutive. Find the remainder when is divided by .
Solution:
Let be the number of ways in which distinct numbers can be selected from the set of the first natural numbers, and let be the number of ways in which distinct numbers, no two of which are consecutive, can be selected from the same set. Then . Because , the problem is reduced to finding .
Consider the natural numbers . If no two of them are consecutive, the numbers , , and are distinct numbers from the interval . Conversely, if are distinct natural numbers from the interval , then , and are from the interval , and no two of them are consecutive. Therefore counting is the same as counting the number of ways of choosing distinct numbers from the set of the first natural numbers. Thus . Hence and the answer is .
The problems on this page are the property of the MAA's American Mathematics Competitions