Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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\end{align*}</cmath>~lopkiloinm | \end{align*}</cmath>~lopkiloinm | ||
==Solution 5(Combinatorics)== | ==Solution 5(Combinatorics)== | ||
− | Suppose you are picking a permutation of <math>2022</math> elements. It ends up being the probability of making every permutation except one permutation so it is <math>1-\frac{1}{2022!}</math>. ~lopkiloinm | + | Suppose you are picking a permutation of <math>2022</math> elements. Suppose that the correct order of the permutation is |
+ | <cmath>x_1,x_2,x_3,\ldots,x_2021,x_2022</cmath> | ||
+ | It ends up being the probability of making every permutation except one permutation so it is <math>1-\frac{1}{2022!}</math>. ~lopkiloinm | ||
== Video Solution == | == Video Solution == |
Revision as of 04:02, 22 November 2022
Contents
Problem
The sum
can be expressed as
, where
and
are positive integers. What is
?
Solution 1
Note that , and therefore this sum is a telescoping sum, which is equivalent to
. Our answer is
.
~mathboy100
Solution 2
We have from canceling a 2022 from
.
This sum clearly telescopes, thus we end up with
. Thus the original equation is equal to
, and
.
.
~not_slay (+ minor LaTeX edit ~TaeKim)
Solution 3 (Induction)
By looking for a pattern, we see that and
, so we can conclude by engineer's induction that the sum in the problem is equal to
, for an answer of
. This can be proven with actual induction as well; we have already established
base cases, so now assume that
for
. For
we get
, completing the proof.
~eibc
Solution 4
Let
~lopkiloinm
Solution 5(Combinatorics)
Suppose you are picking a permutation of elements. Suppose that the correct order of the permutation is
It ends up being the probability of making every permutation except one permutation so it is
. ~lopkiloinm
Video Solution
- Whiz
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.