Difference between revisions of "2012 AMC 12B Problems/Problem 12"
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Consider the 20 term sequence of 0's and 1'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=382</math> so that the answer is <math>\framebox{E}</math>. | Consider the 20 term sequence of 0's and 1'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=382</math> so that the answer is <math>\framebox{E}</math>. | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC12 box|year=2012|ab=B|num-b=11|num-a=13}} | ||
Revision as of 22:20, 12 January 2013
Solution 1
There are 20 Choose 2 selections however, we count these twice therefore
2* 20 C 2 = 380. 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 0's and 1'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 
so that the answer is 
.
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 | |
