https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ender+Wiggin&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-23T18:46:13Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_12&diff=56690 2012 AMC 12B Problems/Problem 12 2013-07-22T14:46:20Z <p>Ender Wiggin: There was a slight typo, of 21 instead of 20.</p> <hr /> <div>How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> There are &lt;math&gt;\binom{20}{2}&lt;/math&gt; selections; however, we count these twice, therefore<br /> <br /> &lt;math&gt;2*\binom{20}{2} = 380&lt;/math&gt;. The wording of the question implies D not E.<br /> <br /> MAA decided to accept both D and E, however.<br /> <br /> ==Solution 2==<br /> <br /> Consider the 20 term sequence of 0's and 1's. Keeping all other terms 1, a sequence of &lt;math&gt;k&gt;0&lt;/math&gt; consecutive 0's can be placed in &lt;math&gt;21-k&lt;/math&gt; 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 &lt;math&gt;20+19+\cdots+1=\binom{21}{2}&lt;/math&gt; strings with consecutive zeros. The same argument shows there are &lt;math&gt;\binom{21}{2}&lt;/math&gt; strings with consecutive 1's. This yields &lt;math&gt;2\binom{21}{2}&lt;/math&gt; 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 &lt;math&gt;01111...&lt;/math&gt;, &lt;math&gt;00111...&lt;/math&gt;, &lt;math&gt;000111...&lt;/math&gt;, ..., &lt;math&gt;000...0001&lt;/math&gt; (of which there are 19) as well as the cases &lt;math&gt;10000...&lt;/math&gt;, &lt;math&gt;11000...&lt;/math&gt;, &lt;math&gt;111000...&lt;/math&gt;, ..., &lt;math&gt;111...110&lt;/math&gt; (of which there are 19 as well). This yields &lt;math&gt;2\binom{21}{2}-2\cdot19=382&lt;/math&gt; so that the answer is &lt;math&gt;\framebox{E}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Ender Wiggin