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

(Created page with "==Problem== How many <math>15</math>-letter arrangements of <math>5</math> A's, <math>5</math> B's, and <math>5</math> C's have no A's in the first <math>5</math> letters, no B's...")
 
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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/0W3VmFp55cM?t=3737
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~ pi_is_3.14
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== Video Solution (Meta-Solving Technique) ==
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https://youtu.be/GmUWIXXf_uk?t=260
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~ pi_is_3.14
  
 
==Solution==
 
==Solution==
{{solution}}
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The answer is <math>\boxed{\textrm{(A)}}</math>.
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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>.
  
 
==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 21:03, 24 January 2021

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/0W3VmFp55cM?t=3737

~ pi_is_3.14

Video Solution (Meta-Solving Technique)

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

~ pi_is_3.14

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 $\sum_{k=0}^{5}\binom{5}{k}^{3}$.

See Also

2003 AMC 12A (ProblemsAnswer KeyResources)
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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