Committee of at least 3 from 20 people

Oct 2009
255
20
St. Louis Area
Given 20 people, how many ways are there to form a committee containing at least three people?
\(\displaystyle \sum_{k=3}^{20}\left(\frac{20}{k}\right)\) is one answer but a shorter calculation would be \(\displaystyle 2^{20}-\left({20\atop 1}\right)-\left({20\atop 2}\right)\).

Does this look correct?
 

Soroban

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May 2006
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Hello, oldguynewstudent!

Given 20 people, how many ways are there to form
a committee containing at least three people?

\(\displaystyle \sum_{k=3}^{20}\left(\frac{20}{k}\right)\) is one answer but a shorter calculation would be \(\displaystyle 2^{20}-\left({20\atop 1}\right)-\left({20\atop 2}\right)\).

Does this look correct? . . . . um, not quite

I assume you mean: .\(\displaystyle \sum^{20}_{k=3}{20\choose k}\)

Also, the alternate way is: .\(\displaystyle 2^{20} - {20\choose0} - {20\choose1} - {20\choose2}\)

 
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Oct 2009
255
20
St. Louis Area
Hello, oldguynewstudent!


I assume you mean: .\(\displaystyle \sum^{20}_{k=3}{20\choose k}\)

Also, the alternate way is: .\(\displaystyle 2^{20} - {20\choose0} - {20\choose1} - {20\choose2}\)
Thanks, yes because you can't have a committee of no people.