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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[[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?
![$\mathrm{(A) \ } 1/3 \qquad \mathrm{(B) \ }3/8 \qquad \mathrm{(C) \ }5/12 \qquad \mathrm{(D) \ } 1/2 \qquad \mathrm{(E) \ }9/16$](http://latex.artofproblemsolving.com/4/8/9/48929a6df67db9f23ea73dd1167b5eb9c1da1e14.png)
Solution
The probability is .