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Difference between revisions of "2003 AMC 12A Problems/Problem 20"

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<math> \textrm{(A)}\ \sum_{k=0}^{5}\binom{5}{k}^{3}\qquad\textrm{(B)}\ 3^{5}\cdot 2^{5}\qquad\textrm{(C)}\ 2^{15}\qquad\textrm{(D)}\ \frac{15!}{(5!)^{3}}\qquad\textrm{(E)}\ 3^{15} </math>
 
<math> \textrm{(A)}\ \sum_{k=0}^{5}\binom{5}{k}^{3}\qquad\textrm{(B)}\ 3^{5}\cdot 2^{5}\qquad\textrm{(C)}\ 2^{15}\qquad\textrm{(D)}\ \frac{15!}{(5!)^{3}}\qquad\textrm{(E)}\ 3^{15} </math>
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== Video Solution ==
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https://youtu.be/3MiGotKnC_U?t=2323
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~ ThePuzzlr
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== Video Solution ==
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https://youtu.be/0W3VmFp55cM?t=3737
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~ Sohil Rathi
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== Video Solution (Meta-Solving Techniques) ==
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https://youtu.be/GmUWIXXf_uk?t=260
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~ Sohil Rathi
  
 
==Solution==
 
==Solution==
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Note that the first five letters must be B's or C's, the next five letters must be C's or A's, and the last five letters must be A's or B's. If there are <math>k</math> B's in the first five letters, then there must be <math>5-k</math> C's in the first five letters, so there must be <math>k</math> C's and <math>5-k</math> A's in the next five letters, and <math>k</math> A's and <math>5-k</math> B's in the last five letters. Therefore the number of each letter in each group of five is determined completely by the number of B's in the first 5 letters, and the number of ways to arrange these 15 letters with this restriction is <math>\binom{5}{k}^3</math> (since there are <math>\binom{5}{k}</math> ways to arrange <math>k</math> B's and <math>5-k</math> C's). Therefore the answer is <math>\sum_{k=0}^{5}\binom{5}{k}^{3}</math>.
 
Note that the first five letters must be B's or C's, the next five letters must be C's or A's, and the last five letters must be A's or B's. If there are <math>k</math> B's in the first five letters, then there must be <math>5-k</math> C's in the first five letters, so there must be <math>k</math> C's and <math>5-k</math> A's in the next five letters, and <math>k</math> A's and <math>5-k</math> B's in the last five letters. Therefore the number of each letter in each group of five is determined completely by the number of B's in the first 5 letters, and the number of ways to arrange these 15 letters with this restriction is <math>\binom{5}{k}^3</math> (since there are <math>\binom{5}{k}</math> ways to arrange <math>k</math> B's and <math>5-k</math> C's). Therefore the answer is <math>\sum_{k=0}^{5}\binom{5}{k}^{3}</math>.
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==Solution 2 (Answer Choices)==
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Note that answer choices B and E have a power of 3 in them, which is impossible since every letter only has 2 options due to the restriction. Therefore, they are automatically eliminated.
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Answer choice C is eliminated due to the fact that not every combination is counted for in 2^15.
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We can show this in a very simple way:
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If all 5 C’s are used in the first 5 letters of the arrangements, then the next 5 letters must be A’s, as B’s are not allowed according to the question. Therefore, there is only one option for these 5 letters, showing a case not accounted for in 2^15.
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Answer choice D is eliminated because although it finds the amount of possible arrangements, it ignores the restriction of the question (A’s cannot be in the first 5 letters, B’s in the second 5, etc.).
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Therefore, the only remaining answer choice is <math>\boxed{\textrm{(A)}}</math>.
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-SwordOfJustice
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 00:12, 12 April 2023

Problem

How many $15$-letter arrangements of $5$ A's, $5$ B's, and $5$ C's have no A's in the first $5$ letters, no B's in the next $5$ letters, and no C's in the last $5$ letters?

$\textrm{(A)}\ \sum_{k=0}^{5}\binom{5}{k}^{3}\qquad\textrm{(B)}\ 3^{5}\cdot 2^{5}\qquad\textrm{(C)}\ 2^{15}\qquad\textrm{(D)}\ \frac{15!}{(5!)^{3}}\qquad\textrm{(E)}\ 3^{15}$

Video Solution

https://youtu.be/3MiGotKnC_U?t=2323

~ ThePuzzlr

Video Solution

https://youtu.be/0W3VmFp55cM?t=3737

~ Sohil Rathi

Video Solution (Meta-Solving Techniques)

https://youtu.be/GmUWIXXf_uk?t=260

~ Sohil Rathi

Solution

The answer is $\boxed{\textrm{(A)}}$.

Note that the first five letters must be B's or C's, the next five letters must be C's or A's, and the last five letters must be A's or B's. If there are $k$ B's in the first five letters, then there must be $5-k$ C's in the first five letters, so there must be $k$ C's and $5-k$ A's in the next five letters, and $k$ A's and $5-k$ B's in the last five letters. Therefore the number of each letter in each group of five is determined completely by the number of B's in the first 5 letters, and the number of ways to arrange these 15 letters with this restriction is $\binom{5}{k}^3$ (since there are $\binom{5}{k}$ ways to arrange $k$ B's and $5-k$ C's). Therefore the answer is $\sum_{k=0}^{5}\binom{5}{k}^{3}$.


Solution 2 (Answer Choices)

Note that answer choices B and E have a power of 3 in them, which is impossible since every letter only has 2 options due to the restriction. Therefore, they are automatically eliminated.


Answer choice C is eliminated due to the fact that not every combination is counted for in 2^15. We can show this in a very simple way:

If all 5 C’s are used in the first 5 letters of the arrangements, then the next 5 letters must be A’s, as B’s are not allowed according to the question. Therefore, there is only one option for these 5 letters, showing a case not accounted for in 2^15.


Answer choice D is eliminated because although it finds the amount of possible arrangements, it ignores the restriction of the question (A’s cannot be in the first 5 letters, B’s in the second 5, etc.).


Therefore, the only remaining answer choice is $\boxed{\textrm{(A)}}$.

-SwordOfJustice

See Also

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

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