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...") |
(→Video Solution (Meta-Solving Techniques)) |
||
| (10 intermediate revisions by 5 users not shown) | |||
| Line 3: | Line 3: | ||
<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> | ||
| + | |||
| + | == 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== | ==Solution== | ||
| − | {{ | + | The answer is <math>\boxed{\textrm{(A)}}</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>. | ||
==See Also== | ==See Also== | ||
| Line 11: | Line 28: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 18:58, 16 January 2023
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/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 
.
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. 
