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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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==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==

Revision as of 13:46, 30 September 2011

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?

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

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 (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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