Difference between revisions of "2003 AMC 12B Problems/Problem 19"
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== Solution == | == Solution == | ||
There are <math>4</math> choices for the first element of <math>S</math>, and for each of these choices there are <math>4!</math> ways to arrange the remaining elements. If the second element must be <math>2</math>, then there are only <math>3</math> choices for the first element and <math>3!</math> ways to arrange the remaining elements. Hence the answer is <math>\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}</math>, and <math>a+b=19 \Rightarrow \mathrm{(E)}</math>. | There are <math>4</math> choices for the first element of <math>S</math>, and for each of these choices there are <math>4!</math> ways to arrange the remaining elements. If the second element must be <math>2</math>, then there are only <math>3</math> choices for the first element and <math>3!</math> ways to arrange the remaining elements. Hence the answer is <math>\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}</math>, and <math>a+b=19 \Rightarrow \mathrm{(E)}</math>. | ||
+ | |||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | There is a <math>\frac {1}{4}</math> chance that the number <math>1</math> is the second term. Let <math>x</math> be the chance that <math>2</math> will be the second term. Since <math>3, 4,</math> and <math>5</math> are in similar situations as <math>2</math>, this becomes <math>\frac {1}{4} + 4x = 1</math> | ||
+ | |||
+ | Solving for <math>x</math>, we find it equals <math>\frac {3}{16}</math>, therefore <math>3 + 16 = 19 \Rightarrow \mathrm{(E)}</math> | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Let's focus on 2, 3, 4, and 5 right now. There are <math>4!</math> ways to arrange these numbers. We can "insert" 1 into the arrangement after the first, second, third, and fourth numbers. There are <math>4! \cdot 4 = 96</math> 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. <math>\frac{\frac{72}{4}}{96} = \frac{3}{16}</math>, therefore <math>3 + 16 = 19 \Rightarrow \mathrm{(E)}</math> | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/IRyWOZQMTV8?t=1215 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
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[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 22:59, 5 May 2024
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.