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 == | ||
+ | https://youtu.be/0W3VmFp55cM?t=3737 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Solution== | ==Solution== |
Revision as of 22:29, 16 January 2021
Contents
Problem
How many -letter arrangements of A's, B's, and C's have no A's in the first letters, no B's in the next letters, and no C's in the last letters?
Video Solution
https://youtu.be/0W3VmFp55cM?t=3737
~ pi_is_3.14
Solution
The answer is .
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 B's in the first five letters, then there must be C's in the first five letters, so there must be C's and A's in the next five letters, and A's and 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 (since there are ways to arrange B's and C's). Therefore the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.