Difference between revisions of "2012 AMC 12B Problems/Problem 12"
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| + | ==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{20}{2}</math> selections; however, we count these twice, therefore | ||
| + | |||
| + | <math>2\cdot\binom{20}{2} = \boxed{\textbf{(D)}\ 380}</math>. 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 <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> | ||
| + | |||
| + | ==Solution 3== | ||
| + | First, we think of ways to make all the <math>1</math>'s consecutive. If there are no consecutive <math>1</math>'s, there are <math>\binom{20}{0}</math> ways to order them. If there is one consecutive <math>1</math>, there are <math>\binom{20}{1}</math> ways to order them. If there are two consecutive <math>1</math>'s, then there are <math>\binom{19}{1}</math> ways to order them (We treat the two <math>1</math>'s like a block, and then order that block with 18 other <math>0</math>'s). Continuing in this fashion, there are <math>\binom{20}{0} + \binom{20}{1} + \binom{19}{1} + \cdots + \binom{1}{1} = 1 + 20 + 19 + \cdots + 2 + 1 = 210 + 1 = 211</math> ways to order consecutive <math>1</math>'s. From symmetry, there are also <math>211</math> ways to order the <math>0</math>'s. Now, from PIE, we subtract out the cases where both the <math>1</math>'s and the <math>0</math>'s are consecutive. We do this because when counting the ways to order the <math>1</math>'s, we counted all of these cases once. Then, we did so again when ordering the <math>0</math>'s. So, to only have all of these cases once, we must subtract them. If <math>1</math> is the leftmost digit, then there are <math>20</math> cases where all the <math>1</math>'s and <math>0</math>'s are consecutive (we basically are choosing how many <math>1</math>'s are consecutive, and there are <math>20</math> possibilities. All other digits become <math>0</math>, which are automatically consecutive since the <math>1</math>'s are consecutive. There are also <math>20</math> cases when <math>0</math> is the left-most digit. Thus, there are a total of <math>211 + 211 - 20 - 20 = \boxed{\textbf{(E)}\ 382}</math>. But, from the way the problem is worded, it somewhat implies that the orderings must include both <math>1</math>'s and <math>0</math>'s, so the answer would then be <math>\boxed{\textbf{(D)}\ 380}</math> after we subtract out the cases where the orderings are either all <math>1</math>'s or all <math>0</math>'s. But, since this is unclear, MAA accepted both <math>\boxed{\textbf{(D}\ 380}</math> and <math>\boxed{\textbf{(E)}\ 382}</math> as acceptable answers. | ||
| + | |||
| + | |||
| + | ===Solution 4=== | ||
| + | |||
| + | We consider two cases, and subtract their overcount. | ||
| + | |||
| + | |||
| + | Case <math>1</math>: Consecutive <math>0</math>s | ||
| + | |||
| + | If we have one consecutive <math>0</math>, then we have <math>20</math> ways. | ||
| + | If we have two consecutive <math>0</math>s, then we have <math>19</math> ways by thinking of the two consecutives as a block. | ||
| + | Continuing this pattern, if we have twenty consecutive <math>0</math>s, then we have only <math>1</math> way. | ||
| + | |||
| + | Therefore, we have | ||
| + | <math>20+19+\cdots+1=\binom{21}{2}</math> ways for this case. | ||
| + | |||
| + | |||
| + | Case <math>2</math>: Consecutive <math>1</math>s | ||
| + | |||
| + | Notice that if we just swap every <math>0</math> to a <math>1</math> in the previous case, we also have a valid arrangement. | ||
| + | Hence, we also have <math>20+19+\cdots+1=\binom{21}{2}</math> ways for this case. | ||
| + | |||
| − | + | Overcount: | |
| + | Notice that we can have BOTH the <math>0</math>s and the <math>1</math>s be consecutive. These are the cases <math>01111...</math>, <math>00111...</math>, <math>000111...</math>, ..., <math>000...0000</math> which gives us <math>20</math> ways being overcounted. If we invert the <math>0</math>s to <math>1</math>s, we similarly have <math>20</math> more ways, hence we need to subtract <math>40</math> from our total count. (Note: this method of overcounting subtracts out the all <math>0</math>s and the all <math>1</math>s case since the problem implies that there needs to be at least one of each) | ||
| − | |||
| − | + | So we have <math>210 + 210 - 40 = 380</math> ways which gives us <math>\boxed{\textbf{(D)}\ 380}</math>. | |
| − | + | ~xHypotenuse | |
| − | |||
== See Also == | == See Also == | ||
{{AMC12 box|year=2012|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2012|ab=B|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:15, 5 May 2024
Problem
How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
Solutions
Solution 1
There are 
selections; however, we count these twice, therefore
. 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 
's and 
'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 
Solution 3
First, we think of ways to make all the 's consecutive. If there are no consecutive 's, there are 
ways to order them. If there is one consecutive , there are 
ways to order them. If there are two consecutive 's, then there are 
ways to order them (We treat the two 's like a block, and then order that block with 18 other 's). Continuing in this fashion, there are 
ways to order consecutive 's. From symmetry, there are also 
ways to order the 's. Now, from PIE, we subtract out the cases where both the 's and the 's are consecutive. We do this because when counting the ways to order the 's, we counted all of these cases once. Then, we did so again when ordering the 's. So, to only have all of these cases once, we must subtract them. If is the leftmost digit, then there are 
cases where all the 's and 's are consecutive (we basically are choosing how many 's are consecutive, and there are possibilities. All other digits become , which are automatically consecutive since the 's are consecutive. There are also cases when is the left-most digit. Thus, there are a total of 
. But, from the way the problem is worded, it somewhat implies that the orderings must include both 's and 's, so the answer would then be 
after we subtract out the cases where the orderings are either all 's or all 's. But, since this is unclear, MAA accepted both 
and 
as acceptable answers.
Solution 4
We consider two cases, and subtract their overcount.
Case : Consecutive s
If we have one consecutive , then we have ways.
If we have two consecutive s, then we have 
ways by thinking of the two consecutives as a block.
Continuing this pattern, if we have twenty consecutive s, then we have only way.
Therefore, we have
ways for this case.
Case 
: Consecutive s
Notice that if we just swap every to a in the previous case, we also have a valid arrangement.
Hence, we also have ways for this case.
Overcount:
Notice that we can have BOTH the s and the s be consecutive. These are the cases , , , ..., 
which gives us ways being overcounted. If we invert the s to s, we similarly have more ways, hence we need to subtract 
from our total count. (Note: this method of overcounting subtracts out the all s and the all s case since the problem implies that there needs to be at least one of each)
So we have 
ways which gives us .
~xHypotenuse
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
