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Difference between revisions of "2022 AMC 10B Problems/Problem 9"

(→‎Solution 5 (Combinatorics))
(→‎Solution 9 (really, really fast cheese))
 
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==Solution 5 (Combinatorics)==
 
==Solution 5 (Combinatorics)==
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Suppose you are sorting a list containing integers between 1 and 2022. The sorted list looks like
+
Let's examine a tuple <math>\sigma</math> containing <math>2022</math> distinct integers. We want to find the probability of the tuple being unsorted.
−
<cmath>1,2,3,\ldots,2021,2022</cmath>
 
−
We want to find the probability of the list being unsorted.
 
  
−
Suppose that we have sorted everything except our last <math>2</math> elements. That is we have
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Suppose that we are looking at the first two items in our tuple. The probability of the first element being greater than the second element is <math>\frac{1}{2!}</math>.
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<cmath>1,2,3,\ldots,2019,2020</cmath>
 
−
We want pick the next element such that it does not equal to <math>2021</math>. There are <math>1</math> ways to choose that, so we add <math>\frac{1}{2!}</math> to the probability.  
 
  
−
Suppose that we have sorted everything except our last <math>3</math> elements. That is we have
+
When we are looking at the first three items in our tuple, the probability of the second element being greater than the third element and the first element less than or equal to the second element is <math>\frac{2}{3!}</math>.
−
<cmath>1,2,3,\ldots,2018,2019</cmath>
+
 
−
We want pick the next element such that it does not equal to <math>2020</math>. There are <math>2</math> ways to choose that, so we add <math>\frac{2}{3!}</math> to the probability.  
+
Similarly, when we are looking at the first four items in our tuple, the probability of the third element being greater than the fourth element, the second element less than or equal to the third element, and the first element less than or equal to the second element is <math>\frac{3}{4!}</math>.
  
 
More specifically,  
 
More specifically,  
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<cmath>\bigsqcup_{n=2}^{2022}\text{unsorted except for last }n=\text{all permutations}\backslash\text{sorted}</cmath>
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<cmath>\begin{align*}
 +
\bigcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\
 +
\bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\
 +
\frac{\#\left(\bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}\right)}{\#\mathcal{S}}&=\frac{\#\left(\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\right)}{\#\mathcal{S}}\\
 +
\frac{\sum_{n=2}^{2022}\frac{2022!}{n!}(n-1)}{2022!}&=\frac{2022!-1}{2022!}\\
 +
\sum_{n=2}^{2022}\frac{n-1}{n!}&=1-\frac{1}{2022!}
 +
\end{align*}</cmath>
  
 
This series ends up being the probability of having the list unsorted and that is of course <math>1-\frac{1}{2022!}</math>
 
This series ends up being the probability of having the list unsorted and that is of course <math>1-\frac{1}{2022!}</math>
 +
 
~lopkiloinm
 
~lopkiloinm
  
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==Solution 8 (Taylor Series)==
 
==Solution 8 (Taylor Series)==
 
We calculate the Taylor series error to be <math>\frac{1}{2023!}</math> and this error happens <math>2023</math> times so it is <math>\frac{2023}{2023!}</math> which is <math>\frac{1}{2022!}</math>
 
We calculate the Taylor series error to be <math>\frac{1}{2023!}</math> and this error happens <math>2023</math> times so it is <math>\frac{2023}{2023!}</math> which is <math>\frac{1}{2022!}</math>
 +
 +
== Solution 9 (really, really fast cheese) ==
 +
Note that <math>\frac{1}{2!} = 1-\frac{1}{2!}</math>, so we guess <math>a=1</math> and <math>b=2022</math>. We get <math>1+2022=\boxed{\textbf{(D) }2023}.</math>
  
 
== Video Solution ==
 
== Video Solution ==

Latest revision as of 12:36, 16 April 2024

Problem

The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$, where $a$ and $b$ are positive integers. What is $a+b$?

$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution 1

Note that $\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$, and therefore this sum is a telescoping sum, which is equivalent to $1 - \frac{1}{2022!}$. Our answer is $1 + 2022 = \boxed{\textbf{(D)}\ 2023}$.

~mathboy100

Solution 2

We add $\frac{1}{2022!}$ to the original expression \[\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\right)+\frac{1}{2022!}=\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2020}{2021!}\right)+\frac{1}{2021!}.\] This sum clearly telescopes, thus we end up with $\left(\frac{1}{2!}+\frac{2}{3!}\right)+\frac{1}{3!}=\frac{2}{2!}=1$. Thus the original expression is equal to $1-\frac{1}{2022!}$, and $1+2022=\boxed{\textbf{(D)}\ 2023}$.

~not_slay (+ minor LaTeX edit ~TaeKim)

Solution 3 (Induction)

By looking for a pattern, we see that $\tfrac{1}{2!} = 1 - \tfrac{1}{2!}$ and $\tfrac{1}{2!} + \tfrac{2}{3!} = \tfrac{5}{6} = 1 - \tfrac{1}{3!}$, so we can conclude by engineer's induction that the sum in the problem is equal to $1 - \tfrac{1}{2022!}$, for an answer of $\boxed{\textbf{(D)}\ 2023}$. This can be proven with actual induction as well; we have already established $2$ base cases, so now assume that $\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n!} = 1 - \tfrac{1}{n!}$ for $n = k$. For $n = k + 1$ we get $\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{1}{n!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{n+1}{(n+1)!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{1}{(n+1)!}$, completing the proof. ~eibc

Solution 4

Let $x=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}.$

Note that \begin{align*} \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=x+1-\frac{1}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)&=1-\frac{1}{2022!}. \end{align*} Therefore, the answer is $1+2022=\boxed{\textbf{(D) }2023}.$

~lopkiloinm

Solution 5 (Combinatorics)

Let's examine a tuple $\sigma$ containing $2022$ distinct integers. We want to find the probability of the tuple being unsorted.

Suppose that we are looking at the first two items in our tuple. The probability of the first element being greater than the second element is $\frac{1}{2!}$.

When we are looking at the first three items in our tuple, the probability of the second element being greater than the third element and the first element less than or equal to the second element is $\frac{2}{3!}$.

Similarly, when we are looking at the first four items in our tuple, the probability of the third element being greater than the fourth element, the second element less than or equal to the third element, and the first element less than or equal to the second element is $\frac{3}{4!}$.

More specifically, \begin{align*} \bigcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\ \bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\ \frac{\#\left(\bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}\right)}{\#\mathcal{S}}&=\frac{\#\left(\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\right)}{\#\mathcal{S}}\\ \frac{\sum_{n=2}^{2022}\frac{2022!}{n!}(n-1)}{2022!}&=\frac{2022!-1}{2022!}\\ \sum_{n=2}^{2022}\frac{n-1}{n!}&=1-\frac{1}{2022!} \end{align*}

This series ends up being the probability of having the list unsorted and that is of course $1-\frac{1}{2022!}$

~lopkiloinm

Solution 6 (Desperate)

Because the fractions get smaller, it is obvious that the answer is less than $1$, so we can safely assume that $a=1$ (this can also be guessed by intuition using similar math problems). Looking at the answer choices, $2018<b<2024$. Because the last term consists of $2022!$ (and the year is $2022$) we can guess that $b=2022$. Adding them yields $1+2022=\boxed{\textbf{(D) }2023}$.

~iluvme and andy_lee

Solution 7 (Partial Fractions)

Knowing that the answer will be in the form $a-\frac{1}{b!}$, we can guess that the sum telescopes. Using partial fractions, we can hope to rewrite $\frac{n-1}{n!}$ as $\frac{A}{(n-1)!}-\frac{B}{n}$. Setting these equal and multiplying by $n!$, we get $n-1=An-B(n-1)!$. Since $An$ is the only term with $n$ with degree $1$, we can conclude that $A=1$. This means that $B=\frac{1}{(n-1)!}$. Substituting, we find that $\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$. This sum clearly telescopes and we obtain $1-\frac{1}{2022!}$. This means that our desired answer is $1+2022=\boxed{\textbf{(D) }2023}.$

~kn07

Solution 8 (Taylor Series)

We calculate the Taylor series error to be $\frac{1}{2023!}$ and this error happens $2023$ times so it is $\frac{2023}{2023!}$ which is $\frac{1}{2022!}$

Solution 9 (really, really fast cheese)

Note that $\frac{1}{2!} = 1-\frac{1}{2!}$, so we guess $a=1$ and $b=2022$. We get $1+2022=\boxed{\textbf{(D) }2023}.$

Video Solution

by Ismail.maths https://www.youtube.com/watch?v=lt34QscjTf4&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj

https://youtu.be/4vdVGYXGvzg

- Whiz

Video Solution by Interstigation

https://youtu.be/_KNR0JV5rdI?t=1102

Video Solution by paixiao

https://www.youtube.com/watch?v=MRwZ4F7Scog

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AMC 10 Problems and Solutions

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