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

(Solution 5(Combinatorics))
(Solution 5(Combinatorics))
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We want to find the probability of picking the permutation in the wrong order. Suppose that we have picked everything to the correct order except our last <math>2</math> elements. That is we have
 
We want to find the probability of picking the permutation in the wrong order. Suppose that we have picked everything to the correct order except our last <math>2</math> elements. That is we have
 
<cmath>x_1,x_2,x_3,\ldots,x_{2019},x_{2020}</cmath>
 
<cmath>x_1,x_2,x_3,\ldots,x_{2019},x_{2020}</cmath>
We want pick the next element such that it does not equal to <math>x_{2021}</math>. There are <math>1</math> ways to choose that, so we add <math>\frac{1}{2!} to the number. Suppose that we have picked everything to the correct order except our last </math>2<math> elements. That is we have
+
We want pick the next element such that it does not equal to <math>x_{2021}</math>. There are <math>1</math> ways to choose that, so we add <math>\frac{1}{2!}</math> to the number. Suppose that we have picked everything to the correct order except our last <math>2</math> elements. That is we have
 
<cmath>x_1,x_2,x_3,\ldots,x_{2018},x_{2019}</cmath>
 
<cmath>x_1,x_2,x_3,\ldots,x_{2018},x_{2019}</cmath>
We want pick the next element such that it does not equal to </math>x_{2020}<math>. There are </math>2<math> ways to choose that, so we add </math>\frac{2}{3!} to the number.  
+
We want pick the next element such that it does not equal to <math>x_{2020}</math>. There are <math>2</math> ways to choose that, so we add <math>\frac{2}{3!}</math> to the number.  
 
~lopkiloinm
 
~lopkiloinm
  

Revision as of 04:08, 22 November 2022

Problem

The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\]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 have $\left(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2021}{2022!}\right)+\frac{1}{2022!}=\left(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2020}{2021!}\right)+\frac{1}{2021!}$ from canceling a 2022 from $\frac{2021+1}{2022!}$. 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 equation is equal to $1-\frac{1}{2022}$, and $1+2022=2023$. $\boxed{\textbf{(D)}\ 2023}$.

~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 $x=\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)$ \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)&=\left(\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\right)\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\right)\\ \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*}~lopkiloinm

Solution 5(Combinatorics)

Suppose you are picking a permutation of $n$ elements. Suppose that the correct order of the permutation is \[x_1,x_2,x_3,\ldots,x_{2021},x_{2022}\] We want to find the probability of picking the permutation in the wrong order. Suppose that we have picked everything to the correct order except our last $2$ elements. That is we have \[x_1,x_2,x_3,\ldots,x_{2019},x_{2020}\] We want pick the next element such that it does not equal to $x_{2021}$. There are $1$ ways to choose that, so we add $\frac{1}{2!}$ to the number. Suppose that we have picked everything to the correct order except our last $2$ elements. That is we have \[x_1,x_2,x_3,\ldots,x_{2018},x_{2019}\] We want pick the next element such that it does not equal to $x_{2020}$. There are $2$ ways to choose that, so we add $\frac{2}{3!}$ to the number. ~lopkiloinm

Video Solution

https://youtu.be/4vdVGYXGvzg

- Whiz

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 10 Problems and Solutions

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