Difference between revisions of "2022 AMC 10A Problems/Problem 24"
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~oh54321 | ~oh54321 | ||
+ | |||
+ | Note: If you are confused, the problem is equivalent to “How many strings of length <math>5</math> formed from the digits <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math> are there such that for each <math>j \in \{1,2,3,4,5\}</math>, at least <math>j</math> of the digits are less than <math>j</math>?” | ||
+ | |||
+ | ~inaccessibles | ||
==Solution 2 (Casework)== | ==Solution 2 (Casework)== | ||
Line 143: | Line 147: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution 4 ( | + | ==Solution 4 (Fake solve, incorrect logic, correct answer by coincidence)== |
+ | |||
+ | The number of strings is <math>(n+1)^{(n-1)}</math> as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose <math>\boxed{\textbf{(E) }1296}</math>. | ||
+ | |||
+ | However, you **CANNOT** prove that fact from this *incorrect* argument: | ||
+ | |||
Let the set of all valid sequences be <math>S</math>. | Let the set of all valid sequences be <math>S</math>. | ||
− | Notice that for any sequence <math>\{a_1,a_2,a_3,a_4,a_5\}</math> in <math>S</math>, the | + | Notice that for any sequence <math>\{a_1,a_2,a_3,a_4,a_5\}</math> in <math>S</math>, the 5 cyclic permutations |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
+ | \{a_1,a_2,a_3,a_4,a_5\}\ | ||
\{a_2,a_3,a_4,a_5,a_1\}\ | \{a_2,a_3,a_4,a_5,a_1\}\ | ||
\{a_3,a_4,a_5,a_1,a_2\}\ | \{a_3,a_4,a_5,a_1,a_2\}\ | ||
Line 154: | Line 164: | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | must also belong in <math>S</math>. However, one must consider the edge case all 5 elements are the same (only <math>\{0,0,0,0,0\}</math>), in which case all sequences listed are equivalent. | + | must also belong in <math>S</math>. However, one must consider the edge case all 5 elements are the same (only <math>\{0,0,0,0,0\}</math>), in which case all sequences listed are equivalent. |
+ | |||
+ | Alas, one must also consider cases like 10101 that only have 3 distinct cyclic permutations, so we cannot get a useful constraint from this view, unless *all* the cases are categorized and counted. | ||
+ | |||
+ | Thus it is NOT implied that <math>\lvert S \rvert \equiv 1 \pmod 5</math>, so it is not justified to choose <math>\boxed{\textbf{(E) }1296}</math> by inspection of answer options. | ||
+ | |||
+ | If you solve <math>S(n)</math> for smaller <math>n</math> by bashing, you can find <math>|S(1)| =1</math>, <math>|S(2)| = 3</math>, <math>|S(3)| =16</math>, <math>|S(4)| =125</math>, | ||
+ | which leads to an Engineer's Induction guess that <math>|S(n)| \equiv 1 \pmod n</math>, or the exact formula <math>(n+1)^{(n-1)}</math> | ||
+ | |||
+ | ~ idea by Tau, logic corrected by oinava | ||
+ | |||
+ | ==Solution 5 (Mod 5 Trick Redeemed)== | ||
+ | Solution 4 tried to observe the answer modulo <math>5</math> to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though: | ||
+ | |||
+ | Notice that for any valid set <math>\{a_1, a_2, a_3, a_4, a_5\}</math>, if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by <math>5</math> (alternatively there is no factor of 5 in the denominator of the multinomial coefficient unless an element is repeated 5 times). Therefore to evaluate the answer modulo <math>5</math>, we only need to look at sets where each element has a multiplicity of at least <math>2</math> (i.e. appears twice or more in the set). | ||
+ | |||
+ | These sets are of the form <math>\{a,a,b,b,b\}</math> and <math>\{a,a,a,a,a\}</math>. The first set can be permuted in <math>\binom{5}{2,3}=10 \equiv 0\pmod{5}</math>, and the second set can be permuted one way, and the only set of the form <math>\{a,a,a,a,a\}</math> is <math>\{0,0,0,0,0\}</math>. Therefore the answer is congruent to <math>1\pmod{5}</math> and you CAN choose <math>\boxed{\textbf{(E) }1296}</math>. | ||
+ | |||
+ | ~SpencerD. | ||
+ | |||
+ | ==Solution 6 (Casework Bash)== | ||
+ | We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string. | ||
+ | 00000 - 1 | ||
+ | |||
+ | 00001 - 5 | ||
+ | |||
+ | 00002 - 5 | ||
+ | |||
+ | 00003 - 5 | ||
+ | |||
+ | 00004 - 5 | ||
+ | |||
+ | 00011 - 10 | ||
+ | |||
+ | 00012 - 20 | ||
+ | |||
+ | 00013 - 20 | ||
+ | |||
+ | 00014 - 20 | ||
+ | |||
+ | 00022 - 10 | ||
+ | |||
+ | 00023 - 20 | ||
+ | |||
+ | 00024 - 20 | ||
+ | |||
+ | 00033 - 10 | ||
+ | |||
+ | 00034 - 20 | ||
+ | |||
+ | 00111 - 10 | ||
+ | |||
+ | 00112 - 30 | ||
+ | |||
+ | 00113 - 30 | ||
+ | |||
+ | 00114 - 30 | ||
+ | |||
+ | 00122 - 30 | ||
+ | |||
+ | 00123 - 60 | ||
+ | |||
+ | 00124 - 60 | ||
+ | |||
+ | 00133 - 30 | ||
+ | |||
+ | 00134 - 60 | ||
+ | |||
+ | 00222 - 10 | ||
+ | |||
+ | 00223 - 30 | ||
+ | |||
+ | 00224 - 30 | ||
+ | |||
+ | 00233 - 30 | ||
+ | |||
+ | 00234 - 60 | ||
+ | |||
+ | 01111 - 5 | ||
+ | |||
+ | 01112 - 20 | ||
+ | |||
+ | 01113 - 20 | ||
+ | |||
+ | 01114 - 20 | ||
+ | |||
+ | 01122 - 30 | ||
+ | |||
+ | 01123 - 60 | ||
+ | |||
+ | 01124 - 60 | ||
+ | |||
+ | 01133 - 30 | ||
+ | |||
+ | 01134 - 60 | ||
+ | |||
+ | 01222 - 20 | ||
+ | |||
+ | 01223 - 60 | ||
+ | |||
+ | 01224 - 60 | ||
+ | |||
+ | 01233 - 60 | ||
+ | |||
+ | 01234 - 120 | ||
+ | |||
+ | |||
+ | Finally, adding all of these up, we obtain <math>1\cdot 1+5\cdot 5+5\cdot 10+10\cdot 20+10\cdot 30 + 10\cdot 60 + 1\cdot 120 = \boxed{\textbf{(E) }1296}</math>. (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5) | ||
+ | |||
+ | The key to organizing all of these cases is to increase the digits one by one, starting from the right and iterating through all the possible cases. | ||
+ | |||
+ | ~ MathHayden | ||
+ | |||
+ | ==Solution 7 (Complementary Counting)== | ||
+ | Let us call the condition where you require at least <math>j</math> digits less than <math>j</math> Rule <math>j</math>. | ||
+ | For a sequence to break Rule <math>1</math>, it must contain no <math>0</math>s. There are <math>4^5=1024</math> ways to do this. | ||
+ | For a sequence to break Rule <math>2</math> but not Rule <math>1</math>, it must contain exactly 1 <math>0</math>. However, this one <math>0</math> contributes toward completing the rule, meaning all other digits must be greater than <math>2</math>. After placing the one <math>0</math> in any of <math>5</math> slots, there are <math>4</math> slots left with <math>3</math> options each, giving a total of <math>5 \cdot 3^4 = 405</math> ways. | ||
+ | For a sequence to break Rule <math>3</math> but neither Rule <math>2</math> or <math>1</math>, it must have two digits less than <math>2</math> (either <math>(0, 0)</math> or <math>(0, 1)</math>) and all other digits must be <math>3</math> or <math>4</math>. There are a total of <math>30</math> ways to arrange the first two digits and <math>2^3</math> ways to arrange the latter three, leaving <math>240</math> ways to do this. | ||
+ | Finally, for a sequence to only break rule <math>4</math>, it must have three digits less than <math>3</math>. They could be <math>(0, 1, 2)</math> (60 arr.), <math>(0, 1, 1); (0, 0, 1); (0, 0, 2)</math> (<math>30</math> arr. each), or <math>(0, 0, 0)</math> (<math>10</math> arr.) This gives <math>160</math> possible arrangements overall. | ||
− | + | There are in total <math>5^5=3125</math> possible arrangements, so by adding all of the invalid states together and subtracting from <math>3125</math>, we get our answer, which is <math>\boxed{\textbf{(E) }1296}</math> | |
+ | ~ Shadowleafy | ||
− | ==Solution | + | ==Video Solution by Math-X (Smart and Simple)== |
+ | https://youtu.be/7yAh4MtJ8a8?si=aBXOzKzoAZwkb4l1&t=9314 | ||
− | + | ~Math-X | |
− | |||
− | |||
==Video Solution== | ==Video Solution== |
Latest revision as of 16:08, 31 October 2024
- 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]- 1 Problem
- 2 Solution 1 (Parking Functions)
- 3 Solution 2 (Casework)
- 4 Solution 3 (Recursive Equations Approach)
- 5 Solution 4 (Fake solve, incorrect logic, correct answer by coincidence)
- 6 Solution 5 (Mod 5 Trick Redeemed)
- 7 Solution 6 (Casework Bash)
- 8 Solution 7 (Complementary Counting)
- 9 Video Solution by Math-X (Smart and Simple)
- 10 Video Solution
- 11 Video Solution
- 12 Video Solution By OmegaLearn using Complementary Counting
- 13 See Also
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
Note: If you are confused, the problem is equivalent to “How many strings of length formed from the digits
,
,
,
,
,
are there such that for each
, at least
of the digits are less than
?”
~inaccessibles
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 (Fake solve, incorrect logic, correct answer by coincidence)
The number of strings is as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose
.
However, you **CANNOT** prove that fact from this *incorrect* argument:
Let the set of all valid sequences be .
Notice that for any sequence
in
, the 5 cyclic permutations
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.
Alas, one must also consider cases like 10101 that only have 3 distinct cyclic permutations, so we cannot get a useful constraint from this view, unless *all* the cases are categorized and counted.
Thus it is NOT implied that , so it is not justified to choose
by inspection of answer options.
If you solve for smaller
by bashing, you can find
,
,
,
,
which leads to an Engineer's Induction guess that
, or the exact formula
~ idea by Tau, logic corrected by oinava
Solution 5 (Mod 5 Trick Redeemed)
Solution 4 tried to observe the answer modulo to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:
Notice that for any valid set , if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by
(alternatively there is no factor of 5 in the denominator of the multinomial coefficient unless an element is repeated 5 times). Therefore to evaluate the answer modulo
, we only need to look at sets where each element has a multiplicity of at least
(i.e. appears twice or more in the set).
These sets are of the form and
. The first set can be permuted in
, and the second set can be permuted one way, and the only set of the form
is
. Therefore the answer is congruent to
and you CAN choose
.
~SpencerD.
Solution 6 (Casework Bash)
We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string. 00000 - 1
00001 - 5
00002 - 5
00003 - 5
00004 - 5
00011 - 10
00012 - 20
00013 - 20
00014 - 20
00022 - 10
00023 - 20
00024 - 20
00033 - 10
00034 - 20
00111 - 10
00112 - 30
00113 - 30
00114 - 30
00122 - 30
00123 - 60
00124 - 60
00133 - 30
00134 - 60
00222 - 10
00223 - 30
00224 - 30
00233 - 30
00234 - 60
01111 - 5
01112 - 20
01113 - 20
01114 - 20
01122 - 30
01123 - 60
01124 - 60
01133 - 30
01134 - 60
01222 - 20
01223 - 60
01224 - 60
01233 - 60
01234 - 120
Finally, adding all of these up, we obtain . (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5)
The key to organizing all of these cases is to increase the digits one by one, starting from the right and iterating through all the possible cases.
~ MathHayden
Solution 7 (Complementary Counting)
Let us call the condition where you require at least digits less than
Rule
.
For a sequence to break Rule
, it must contain no
s. There are
ways to do this.
For a sequence to break Rule
but not Rule
, it must contain exactly 1
. However, this one
contributes toward completing the rule, meaning all other digits must be greater than
. After placing the one
in any of
slots, there are
slots left with
options each, giving a total of
ways.
For a sequence to break Rule
but neither Rule
or
, it must have two digits less than
(either
or
) and all other digits must be
or
. There are a total of
ways to arrange the first two digits and
ways to arrange the latter three, leaving
ways to do this.
Finally, for a sequence to only break rule
, it must have three digits less than
. They could be
(60 arr.),
(
arr. each), or
(
arr.) This gives
possible arrangements overall.
There are in total possible arrangements, so by adding all of the invalid states together and subtracting from
, we get our answer, which is
~ Shadowleafy
Video Solution by Math-X (Smart and Simple)
https://youtu.be/7yAh4MtJ8a8?si=aBXOzKzoAZwkb4l1&t=9314
~Math-X
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.