2012 AMC 12B Problems/Problem 12

Revision as of 22:00, 11 March 2012 by Williamhu888 (talk | contribs) (Solution 1)

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 $k>0$ consecutive 0's can be placed in $21-k$ 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 $20+19+\cdots+1=\binom{21}{2}$ strings with consecutive zeros. The same argument shows there are $\binom{21}{2}$ strings with consecutive 1's. This yields $2\binom{21}{2}$ 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 $01111...$, $00111...$, $000111...$, ..., $000...0001$ (of which there are 19) as well as the cases $10000...$, $11000...$, $111000...$, ..., $111...110$ (of which there are 19 as well). This yields $2\binom{21}{2}-2\cdot19=382$ so that the answer is $\framebox{E}$.

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