Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 11"
m (→Solution) |
|||
| Line 7: | Line 7: | ||
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>. | ||
| + | {{solution}} | ||
---- | ---- | ||
Revision as of 17:18, 17 August 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 
.
This problem needs a solution. If you have a solution for it, please help us out by adding it.
