Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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==Solution 5 (Combinatorics)== | ==Solution 5 (Combinatorics)== | ||
− | Suppose you are picking a permutation of <math>2022</math> elements. Suppose that the correct order of the permutation is | + | Suppose you are picking a permutation of <math>2022</math> distinct elements. Suppose that the correct order of the permutation is |
<cmath>x_1,x_2,x_3,\ldots,x_{2021},x_{2022}</cmath> | <cmath>x_1,x_2,x_3,\ldots,x_{2021},x_{2022}</cmath> | ||
We want to find the probability of picking the permutation in the wrong order. | We want to find the probability of picking the permutation in the wrong order. |
Revision as of 22:50, 25 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
Note that
Therefore, the answer is
~lopkiloinm
Solution 5 (Combinatorics)
Suppose you are picking a permutation of distinct elements. Suppose that the correct order of the permutation is
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 elements. That is we have
We want pick the next element such that it does not equal to
. There are
ways to choose that, so we add
to the probability.
Suppose that we have picked everything to the correct order except our last elements. That is we have
We want pick the next element such that it does not equal to
. There are
ways to choose that, so we add
to the probability.
This series ends up being the probability of making a permutation in the wrong order and that is of course
~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.