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Japan competition problems, 2007

Combinatorics Difficulty 4.6 AIME Find the answer Japan

We have 15 cards numbered 11, 22, …\dots, 1515. How many ways are there to choose some (at least 1) cards so that all numbers on these cards are larger than or equal to the number of cards chosen?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Consider a general problem with cards 11, 22, …\dots, nn. Let FnF_n be the number of choices when there are nn cards. F1=1F_1 = 1 and F2=2F_2 = 2 are trivial. Let k≥3k \ge 3. We will consider FkF_k. If card kk is not chosen, the number of ways is trivially Fk−1F_{k-1}.

Consider the case where card kk is chosen. If any card other than kk is chosen, we cannot choose 11. And if we remove the card kk and decrease the numbers on other cards by 11, we get a proper choice from cards 11, …\dots, k−2k-2. On the other hand, if we have a proper choice from 11, …\dots, k−2k-2, we get a choice with kk and some other cards among 11, …\dots, k−1k-1. If no card other than kk is chosen, there is trivially only 11 way. Therefore, the number of proper choices with card kk is Fk−2+1F_{k-2} + 1.

So we get a relation Fk=Fk−1+Fk−2+1F_k = F_{k-1} + F_{k-2} + 1. Calculate the sequence by this relation we get F15=1596F_{15} = 1596.

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