2003 AMC 12B Problems/Problem 19
Problem
Let be the set of permutations of the sequence for which the first term is not . A permutation is chosen randomly from . The probability that the second term is , in lowest terms, is . What is ?
Solution
There are choices for the first element of , and for each of these choices there are ways to arrange the remaining elements. If the second element must be , then there are only choices for the first element and ways to arrange the remaining elements. Hence the answer is , and .
Solution 2
There is a chance that the number is the second term. Let be the chance that will be the second term. Since and are in similar situations as , this becomes
Solving for , we find it equals , therefore
Video Solution by OmegaLearn
https://youtu.be/IRyWOZQMTV8?t=1215
~ pi_is_3.14
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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All AMC 12 Problems and Solutions |
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