Answer: Complete Solution: At least 677 people must be in the school. There are 26 letters in the English alphabet. So there are 26 different possibilities for the first initial. Consider all the possible pairs of two initials. For example, suppose a person has the first initial A. Then the pair of initials could be AA, AB, AC, ., AZ. There are 26 different possibilities. If the first initial is B, the pair of initials could be BA, BB, BC, ., BZ. Again there are 26 different pairs. Continuing in this way and since there are 26 possible first initials, each of which could be paired with 26 last initials, there are 26 × 26, or 676 possible different pairs of initials. If there were 677 people, at least two of them must have the same pair of initials. Alternate Solution: Another way to begin this problem is to think about a situation involving smaller numbers. Suppose you have 11 items (that cannot be broken) to give to 10 people. This means that 1 person will get 2 of the items. The same reasoning can be used to solve the challenge.
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