Difference between revisions of "2015 AMC 10A Problems/Problem 10"
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math> | <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | The first thing one would want to do is | + | The first thing one would want to do is place a possible letter that works and then stem off of it. For example, if we start with an <math>a</math>, we can only place a <math>c</math> or <math>d</math> next to it. Unfortunately, after that step, we can't do too much, since: |
<math>acbd</math> is not allowed because of the <math>cb</math>, and <math>acdb</math> is not allowed because of the <math>cd</math>. | <math>acbd</math> is not allowed because of the <math>cb</math>, and <math>acdb</math> is not allowed because of the <math>cd</math>. | ||
− | We get the same problem if we start with a <math>d</math>, since a <math>b</math> will have to end up in the middle, adjacent to an <math>a</math> or <math>c</math> | + | We get the same problem if we start with a <math>d</math>, since a <math>b</math> will have to end up in the middle, causing it to be adjacent to an <math>a</math> or <math>c</math>. |
If we start with a <math>b</math>, the next letter would have to be a <math>d</math>, and since we can put an <math>a</math> next to it and then a <math>c</math> after that, this configuration works. The same approach applies if we start with a <math>c</math>. | If we start with a <math>b</math>, the next letter would have to be a <math>d</math>, and since we can put an <math>a</math> next to it and then a <math>c</math> after that, this configuration works. The same approach applies if we start with a <math>c</math>. | ||
− | So the solution must be <math>1 + 1 = \boxed{\ | + | So the solution must be the two solutions that were allowed, one starting from a <math>b</math> and the other with a <math>c</math>, giving us: |
+ | |||
+ | <cmath>1 + 1 = \boxed{\textbf{(C)}\ 2}.</cmath> | ||
+ | |||
+ | |||
+ | ==Solution 2 (Casework)== | ||
+ | |||
+ | '''Case 1: the first letter is A''' | ||
+ | |||
+ | ''Subcase 1: the second letter is C'' | ||
+ | |||
+ | The next letter must either be B or D, both of which do not satisfy the conditions. | ||
+ | |||
+ | ''Subcase 2: the second letter is D'' | ||
+ | |||
+ | The third letter is forced to be B and the fourth is forced to be C, but this doesn't work because B and C are next to each other. | ||
+ | |||
+ | '''Case 2: the first letter is B''' | ||
+ | |||
+ | The next letter is forced to be D, the third letter forced to be A and the last forced to be C. This works. | ||
+ | |||
+ | '''Case 3: the first letter is C''' | ||
+ | |||
+ | The next letter is forced to be A, the third letter is D and the last letter is B. This works. | ||
+ | |||
+ | '''Case 4: the first letter is D''' | ||
+ | |||
+ | ''Subcase 1: the second letter is A'' | ||
+ | |||
+ | The third letter has to be C and the fourth has to be B but this doesn't work because B and C are next to each other. | ||
+ | |||
+ | ''Subcase 2: the second letter is B'' | ||
+ | |||
+ | The third letter cannot be A or C, so this doesn't work. | ||
+ | |||
+ | Summing the cases, there are two that work: <math>BDAC</math> and <math>CADB</math> <math>\Longrightarrow \boxed{\textbf{(C)}\ 2}</math>. | ||
+ | |||
+ | ~JH. L | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/hTcv8lbvs6o | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/0W3VmFp55cM?t=1055 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/3MiGotKnC_U?t=1509 | ||
+ | |||
+ | ~ ThePuzzlr | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/8sTQIX4YJ6s | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2015|ab=A|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] |
Latest revision as of 18:17, 10 August 2023
Contents
[hide]Problem
How many rearrangements of are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either or .
Solution 1
The first thing one would want to do is place a possible letter that works and then stem off of it. For example, if we start with an , we can only place a or next to it. Unfortunately, after that step, we can't do too much, since:
is not allowed because of the , and is not allowed because of the .
We get the same problem if we start with a , since a will have to end up in the middle, causing it to be adjacent to an or .
If we start with a , the next letter would have to be a , and since we can put an next to it and then a after that, this configuration works. The same approach applies if we start with a .
So the solution must be the two solutions that were allowed, one starting from a and the other with a , giving us:
Solution 2 (Casework)
Case 1: the first letter is A
Subcase 1: the second letter is C
The next letter must either be B or D, both of which do not satisfy the conditions.
Subcase 2: the second letter is D
The third letter is forced to be B and the fourth is forced to be C, but this doesn't work because B and C are next to each other.
Case 2: the first letter is B
The next letter is forced to be D, the third letter forced to be A and the last forced to be C. This works.
Case 3: the first letter is C
The next letter is forced to be A, the third letter is D and the last letter is B. This works.
Case 4: the first letter is D
Subcase 1: the second letter is A
The third letter has to be C and the fourth has to be B but this doesn't work because B and C are next to each other.
Subcase 2: the second letter is B
The third letter cannot be A or C, so this doesn't work.
Summing the cases, there are two that work: and .
~JH. L
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://youtu.be/0W3VmFp55cM?t=1055
~ pi_is_3.14
Video Solution
https://youtu.be/3MiGotKnC_U?t=1509
~ ThePuzzlr
Video Solution
~savannahsolver
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.