Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 12B Problems/Problem 19"

(sol)
 
 
(5 intermediate revisions by 4 users not shown)
Line 10: Line 10:
 
== 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>
 +
 +
== Solution 4 ==
 +
 +
The probability could be written as a fraction where <math>\frac {\text{number of viable solutions in S}}{\text{total permutations in S}}</math>.
 +
 +
The total number of permutations is essentially <math>5!</math>, and the number of permutations where <math>1</math> is the first number is <math>4!</math>, therefore the number of permutations in <math>S</math> is <math>5!-4!</math>.
 +
 +
For the first number in the desired solution, there are 3 options (<math>3</math>, <math>4</math>, <math>5</math>).
 +
For the second number in the desired solution, there is only one option (<math>2</math>).
 +
For the third number in the desired solution, there are 3 options (<math>1</math>, 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, <math>3 \cdot 1 \cdot 3 \cdot 2 \cdot 1</math> will be the number of desired outcomes.
 +
 +
Finally, <math>\frac {3 \cdot 1 \cdot 3 \cdot 2 \cdot 1}{5!-4!}</math> equates to <math>\frac {18}{96}</math>, which is <math>\frac {3}{16}</math>.
 +
 +
The answer is then <math>3+16=\boxed{\textbf{(E)}\ 19}</math>.
 +
 +
~ Tyrone12345
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/IRyWOZQMTV8?t=1215
 +
 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==
Line 15: Line 53:
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:28, 15 August 2024

Problem

Let $S$ be the set of permutations of the sequence $1,2,3,4,5$ for which the first term is not $1$. A permutation is chosen randomly from $S$. The probability that the second term is $2$, in lowest terms, is $a/b$. What is $a+b$?

$\mathrm{(A)}\ 5 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 16 \qquad\mathrm{(E)}\ 19$

Solution

There are $4$ choices for the first element of $S$, and for each of these choices there are $4!$ ways to arrange the remaining elements. If the second element must be $2$, then there are only $3$ choices for the first element and $3!$ ways to arrange the remaining elements. Hence the answer is $\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}$, and $a+b=19 \Rightarrow \mathrm{(E)}$.


Solution 2

There is a $\frac {1}{4}$ chance that the number $1$ is the second term. Let $x$ be the chance that $2$ will be the second term. Since $3, 4,$ and $5$ are in similar situations as $2$, this becomes $\frac {1}{4} + 4x = 1$

Solving for $x$, we find it equals $\frac {3}{16}$, therefore $3 + 16 = 19 \Rightarrow \mathrm{(E)}$

Solution 3

Let's focus on 2, 3, 4, and 5 right now. There are $4!$ ways to arrange these numbers. We can "insert" 1 into the arrangement after the first, second, third, and fourth numbers. There are $4! \cdot 4 = 96$ 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. $\frac{\frac{72}{4}}{96} = \frac{3}{16}$, therefore $3 + 16 = 19 \Rightarrow \mathrm{(E)}$

Solution 4

The probability could be written as a fraction where $\frac {\text{number of viable solutions in S}}{\text{total permutations in S}}$.

The total number of permutations is essentially $5!$, and the number of permutations where $1$ is the first number is $4!$, therefore the number of permutations in $S$ is $5!-4!$.

For the first number in the desired solution, there are 3 options ($3$, $4$, $5$). For the second number in the desired solution, there is only one option ($2$). For the third number in the desired solution, there are 3 options ($1$, 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, $3 \cdot 1 \cdot 3 \cdot 2 \cdot 1$ will be the number of desired outcomes.

Finally, $\frac {3 \cdot 1 \cdot 3 \cdot 2 \cdot 1}{5!-4!}$ equates to $\frac {18}{96}$, which is $\frac {3}{16}$.

The answer is then $3+16=\boxed{\textbf{(E)}\ 19}$.

~ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12B_Problems/Problem_19&oldid=225061"