2003 AMC 12B Problems/Problem 19
Contents
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
Solution 4
The probability could be written as a fraction where .
The total number of permutations is essentially , and the number of permutations where is the first number is , therefore the number of permutations in is .
For the first number in the desired solution, there are 3 options (, , ). For the second number in the desired solution, there is only one option (). For the third number in the desired solution, there are 3 options (, and the numbers not used in the first digit). For the fourth number in the desired solution, there are 2 options (numbers not used in the third digit). For the last number in the desired solution, there is only one option (number not used in the fourth digit).
Therefore, will be the number of desired outcomes.
Finally, equates to , which is .
The answer is then .
~ Tyrone12345
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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