# Difference between revisions of "2012 AMC 12B Problems/Problem 12"

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2* 20 C 2 = 380. The wording of the question implies D not E. | 2* 20 C 2 = 380. The wording of the question implies D not E. | ||

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+ | MAA decided to accept both D and E, however. | ||

==Solution 2== | ==Solution 2== | ||

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>. |

## Revision as of 22:00, 11 March 2012

## 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 .