Difference between revisions of "2022 AMC 10A Problems/Problem 24"
Pratimakarpe (talk | contribs) (→Solution 1 (Parking Functions)) |
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | Solution 4 - Casework on Number of 0s (Shiva Kannan) | ||
+ | |||
+ | Case <math>1</math>: String contains <math>5</math> <math>0s</math> | ||
+ | The only string that works is <math>00000</math> which yields <math>1</math> possible string that satisfies the given conditions | ||
+ | Case <math>2</math>: | ||
==Solution 4 (Answer Choices)== | ==Solution 4 (Answer Choices)== |
Revision as of 20:55, 11 January 2023
- The following problem is from both the 2022 AMC 10A #24 and 2022 AMC 12A #24, so both problems redirect to this page.
Contents
[hide]Problem
How many strings of length formed from the digits
,
,
,
,
are there such that for each
, at least
of the digits are less than
? (For example,
satisfies this condition
because it contains at least
digit less than
, at least
digits less than
, at least
digits less
than
, and at least
digits less than
. The string
does not satisfy the condition because it
does not contain at least
digits less than
.)
Solution 1 (Parking Functions)
For some , let there be
parking spaces counterclockwise in a circle. Consider a string of
integers
each between
and
, and let
cars come into this circle so that the
th car tries to park at spot
, but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of numbers between
and
that contain at least
integers
for
are exactly the set of strings that leave spot
empty. Also note for any string
, we can add
to each
(mod
) to shift the empty spot counterclockwise, meaning for each string there exists exactly one
with
so that
leaves spot
empty. This gives there are
such strings.
Plugging in gives
such strings.
~oh54321
Solution 2 (Casework)
Note that a valid string must have at least one
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
- The string has
different digit.
- The string has
different digits.
- The string has
different digits.
- The string has
different digits.
- The string has
different digits.
The only possibility is
There is string in this case.
We have the following table:
There are
strings in this case.
We have the following table:
There are
strings in this case.
We have the following table:
There are
strings in this case.
There are strings in this case.
Together, the answer is
~MRENTHUSIASM
Solution 3 (Recursive Equations Approach)
Denote by the number of
-digit strings formed by using numbers
, where for each
, at least
of the digits are less than
.
We have the following recursive equation:
and the boundary condition
for any
.
By solving this recursive equation, for and
, we get
For and
, we get
For and
, we get
For and
, we get
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4 - Casework on Number of 0s (Shiva Kannan)
Case : String contains
The only string that works is
which yields
possible string that satisfies the given conditions
Case
:
Solution 4 (Answer Choices)
Let the set of all valid sequences be .
Notice that for any sequence
in
, the sequences
must also belong in
. However, one must consider the edge case all 5 elements are the same (only
), in which case all sequences listed are equivalent. Then
, which yields
by inspection.
~Tau
Video Solution
~MathProblemSolvingSkills.com
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution By OmegaLearn using Complementary Counting
~ pi_is_3.14
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.