2003 AMC 12A Problems/Problem 20

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

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

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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2003 AMC 12A Problems/Problem 20

Contents

Problem

Video Solution

Video Solution

Video Solution (Meta-Solving Techniques)

Solution

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