Difference between revisions of "2012 AMC 12B Problems/Problem 12"
(→Solution 2) |
(→Solution 2) |
||
Line 17: | Line 17: | ||
===Solution 2=== | ===Solution 2=== | ||
− | Consider the 20 term sequence of <math>0</math>'s and <math>1</math>'s. Keeping all other terms 1, a sequence of <math>k>0</math> consecutive 0's can be placed in <math>21-k</math> locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are <math>20+19+\cdots+1=\binom{21}{2}</math> strings with consecutive zeros. The same argument shows there are <math>\binom{21}{2}</math> strings with consecutive 1's. This yields <math>2\binom{21}{2}</math> strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases <math>01111...</math>, <math>00111...</math>, <math>000111...</math>, ..., <math>000...0001</math> (of which there are 19) as well as the cases <math>10000...</math>, <math>11000...</math>, <math>111000...</math>, ..., <math>111...110</math> (of which there are 19 as well). This yields <math>2\binom{21}{2}-2\cdot19=\textbf{(E)}\ 382</math> | + | Consider the 20 term sequence of <math>0</math>'s and <math>1</math>'s. Keeping all other terms 1, a sequence of <math>k>0</math> consecutive 0's can be placed in <math>21-k</math> locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are <math>20+19+\cdots+1=\binom{21}{2}</math> strings with consecutive zeros. The same argument shows there are <math>\binom{21}{2}</math> strings with consecutive 1's. This yields <math>2\binom{21}{2}</math> strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases <math>01111...</math>, <math>00111...</math>, <math>000111...</math>, ..., <math>000...0001</math> (of which there are 19) as well as the cases <math>10000...</math>, <math>11000...</math>, <math>111000...</math>, ..., <math>111...110</math> (of which there are 19 as well). This yields <math>2\binom{21}{2}-2\cdot19=\boxed{\textbf{(E)}\ 382}</math> |
== See Also == | == See Also == |
Revision as of 20:41, 10 October 2020
Problem
How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
Solutions
Solution 1
There are selections; however, we count these twice, therefore
. The wording of the question implies D not E.
MAA decided to accept both D and E, however.
Solution 2
Consider the 20 term sequence of 's and 's. Keeping all other terms 1, a sequence of consecutive 0's can be placed in locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are strings with consecutive zeros. The same argument shows there are strings with consecutive 1's. This yields strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases , , , ..., (of which there are 19) as well as the cases , , , ..., (of which there are 19 as well). This yields
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.