Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 11"
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== Problem == | == 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? | ||
| − | <center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ } </math></center> | + | <center><math> \mathrm{(A) \ } 1/3 \qquad \mathrm{(B) \ }3/8 \qquad \mathrm{(C) \ }5/12 \qquad \mathrm{(D) \ } 1/2 \qquad \mathrm{(E) \ }9/16 </math></center> |
== Solution == | == Solution == | ||
| + | The probability is <math>\frac{3!+2!+1!}{4!}=\frac{3}{8}</math>. | ||
== See also == | == See also == | ||
* [[University of South Carolina High School Math Contest/1993 Exam]] | * [[University of South Carolina High School Math Contest/1993 Exam]] | ||
| + | |||
| + | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 20:33, 22 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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