Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 11"
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The probability is <math>\frac{3!+2!+1!}{4!}=\frac{3}{8}</math>. | The probability is <math>\frac{3!+2!+1!}{4!}=\frac{3}{8}</math>. | ||
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| − | * [[University of South Carolina High School Math Contest/1993 Exam]] | + | |
| + | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 10|Previous Problem]] | ||
| + | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 12|Next Problem]] | ||
| + | * [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 12:23, 23 July 2006
Problem
Suppose that 4 cards labeled 1 to 4 are placed randomly into 4 boxes also labeled 1 to 4, one card per box. What is the probability that no card gets placed into a box having the same label as the card?
Solution
The probability is 
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