# Thread: Committee of at least 3 from 20 people

1. ## Committee of at least 3 from 20 people

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?

2. 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: .$\displaystyle 2^{20} - {20\choose0} - {20\choose1} - {20\choose2}$

3. Originally Posted by Soroban
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.