Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 15, 2011"

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{{:AoPSWiki:Problem of the Day/July 15, 2011}}
 
{{:AoPSWiki:Problem of the Day/July 15, 2011}}
 
==Solution==
 
==Solution==
There are <math>\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}</math> ways to satisfy the baby storks, out of a total of <math>2^10</math> ways to catch the fish. Using the fact that <math>\binom{n}{k}=\binom{n}{n-k}</math>, we can quickly evaluate the probability as <math>\frac{270+120+45+10+1}{1024}=\frac{446}{1024}=\boxed{\frac{223}{512}}</math>.
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There are <math>\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}</math> ways to satisfy the baby storks, out of a total of <math>2^{10}</math> ways to catch the fish. Using the fact that <math>\binom{n}{k}=\binom{n}{n-k}</math>, we can quickly evaluate the probability as <math>\frac{270+120+45+10+1}{1024}=\frac{446}{1024}=\boxed{\frac{223}{512}}</math>.

Revision as of 03:52, 18 August 2011

Problem

AoPSWiki:Problem of the Day/July 15, 2011

Solution

There are $\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}$ ways to satisfy the baby storks, out of a total of $2^{10}$ ways to catch the fish. Using the fact that $\binom{n}{k}=\binom{n}{n-k}$, we can quickly evaluate the probability as $\frac{270+120+45+10+1}{1024}=\frac{446}{1024}=\boxed{\frac{223}{512}}$.

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