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
Solution 3
Let's focus on 2, 3, 4, and 5 right now. There are ways to arrange these numbers. We can "insert" 1 into the arrangement after the first, second, third, and fourth numbers. There are ways to do this.
In 24 of those ways, 1 is in the second position, and in the remaining 72 sequences, 2, 3, 4, and 5 occur in the second position the same number of times. , 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 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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