2012 AMC 12B Problems/Problem 12


How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?

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


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

Solution 3

First, we think of ways to make all the $1$'s consecutive. If there are no consecutive $1$'s, there are $\binom{20}{0}$ ways to order them. If there is one consecutive $1$, there are $\binom{20}{1}$ ways to order them. If there are two consecutive $1$'s, then there are $\binom{19}{1}$ ways to order them (We treat the two $1$'s like a block, and then order that block with 18 other $0$'s). Continuing in this fashion, there are $\binom{20}{0} + \binom{20}{1} + \binom{19}{1} + \cdots + \binom{1}{1} = 1 + 20 + 19 + \cdots + 2 + 1 = 210 + 1 = 211$ ways to order consecutive $1$'s. From symmetry, there are also $211$ ways to order the $0$'s. Now, from PIE, we subtract out the cases where both the $1$'s and the $0$'s are consecutive. We do this because when counting the ways to order the $1$'s, we counted all of these cases once. Then, we did so again when ordering the $0$'s. So, to only have all of these cases once, we must subtract them. If $1$ is the leftmost digit, then there are $20$ cases where all the $1$'s and $0$'s are consecutive (we basically are choosing how many $1$'s are consecutive, and there are $20$ possibilities. All other digits become $0$, which are automatically consecutive since the $1$'s are consecutive. There are also $20$ cases when $0$ is the left-most digit. Thus, there are a total of $211 + 211 - 20 - 20 = \boxed{\textbf{(E)}\ 382}$. But, from the way the problem is worded, it somewhat implies that the orderings must include both $1$'s and $0$'s, so the answer would then be $\boxed{\textbf{(D)}\ 380}$ after we subtract out the cases where the orderings are either all $1$'s or all $0$'s. But, since this is unclear, MAA accepted both $\boxed{\textbf{(D}\ 380}$ and $\boxed{\textbf{(E)}\ 382}$ as acceptable answers.

Solution 4

We consider two cases, and subtract their overcount.

Case $1$: Consecutive $0$s

If we have one consecutive $0$, then we have $20$ ways. If we have two consecutive $0$s, then we have $19$ ways by thinking of the two consecutives as a block. Continuing this pattern, if we have twenty consecutive $0$s, then we have only $1$ way.

Therefore, we have $20+19+\cdots+1=\binom{21}{2}$ ways for this case.

Case $2$: Consecutive $1$s

Notice that if we just swap every $0$ to a $1$ in the previous case, we also have a valid arrangement. Hence, we also have $20+19+\cdots+1=\binom{21}{2}$ ways for this case.

Overcount: Notice that we can have BOTH the $0$s and the $1$s be consecutive. These are the cases $01111...$, $00111...$, $000111...$, ..., $000...0000$ which gives us $20$ ways being overcounted. If we invert the $0$s to $1$s, we similarly have $20$ more ways, hence we need to subtract $40$ from our total count. (Note: this method of overcounting subtracts out the all $0$s and the all $1$s case since the problem implies that there needs to be at least one of each)

So we have $210 + 210 - 40 = 380$ ways which gives us $\boxed{\textbf{(D)}\ 380}$.


See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 12 Problems and Solutions

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