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

m
(Solution 1)
 
(9 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
 
How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
 
How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
  
 
<math>\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382</math>
 
<math>\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382</math>
  
==Solution 1==
+
==Solutions==
 +
 
 +
===Solution 1===
  
There are <math>\binom{21}{2}</math> selections; however, we count these twice, therefore
+
There are <math>\binom{20}{2}</math> selections; however, we count these twice, therefore
  
<math>2*\binom{21}{2} = 380</math>. The wording of the question implies D not E.
+
<math>2\cdot\binom{20}{2} = \boxed{\textbf{(D)}\ 380}</math>. The wording of the question implies D, not E.
  
MAA decided to accept both D and E, however.
+
However, MAA decided to accept both D and E.
  
==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 <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 ==

Latest revision as of 20:43, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS