We have 15 cards numbered , , , . 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?
, 2007
Solution
Consider a general problem with cards , , , . Let be the number of choices when there are cards. and are trivial. Let . We will consider . If card is not chosen, the number of ways is trivially .
Consider the case where card is chosen. If any card other than is chosen, we cannot choose . And if we remove the card and decrease the numbers on other cards by , we get a proper choice from cards , , . On the other hand, if we have a proper choice from , , , we get a choice with and some other cards among , , . If no card other than is chosen, there is trivially only way. Therefore, the number of proper choices with card is .
So we get a relation . Calculate the sequence by this relation we get .
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