Problem 12 · AMC 8 Stretch
Core
Counting & Probability
and-process-multiplypigeonholelogical-reasoning
Now there are 6 letters and 4 mailboxes. In how many ways can the letters be placed if (a) each letter must go in a different mailbox; (b) any number of letters may share a mailbox?
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Answer: (a) 0 (impossible); (b) 4096
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Hint 1 of 3
Use the same place-one-letter-at-a-time idea, but now there are more letters than boxes.
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Hint 2 of 3
Part (a) wants all 6 letters in different mailboxes — but there are only 4 boxes. Is that even possible?
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Hint 3 of 3
If 6 letters must each be in a separate box but there are only 4 boxes, two letters are forced to share. So (a) has 0 ways. For (b), each of the 6 letters has 4 choices.
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Approach: Pigeonhole for (a); AND process for (b)
- (a) Different mailboxes: you would need at least 6 boxes to put 6 letters in different boxes, but there are only 4. By the pigeonhole idea (more letters than boxes forces a box to hold two), it is impossible — 0 ways.
- (b) No restriction: place the 6 letters one at a time, each with 4 choices, so \(4^6 = 4096\).
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