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

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There are 20 Choose 2 selections however, we count these twice therefore
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==Problem==
  
2* 20 C 2 = 380. The wording of the question implies D not E.
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How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
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<math>\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382</math>
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==Solutions==
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===Solution 1===
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There are <math>\binom{20}{2}</math> selections; however, we count these twice, therefore
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<math>2\cdot\binom{20}{2} = \boxed{\textbf{(D)}\ 380}</math>. The wording of the question implies D, not E.
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However, MAA decided to accept both D and E.
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===Solution 2===
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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>
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== See Also ==
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{{AMC12 box|year=2012|ab=B|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 21:43, 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?

$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$

Solutions

Solution 1

There are $\binom{20}{2}$ selections; however, we count these twice, therefore

$2\cdot\binom{20}{2} = \boxed{\textbf{(D)}\ 380}$. The wording of the question implies D, not E.

However, MAA decided to accept both D and E.

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 $2\binom{21}{2}-2\cdot19=\boxed{\textbf{(E)}\ 382}$

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
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

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