# Committee of at least 3 from 20 people

#### 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?

#### Soroban

MHF Hall of Honor
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: .$$\displaystyle 2^{20} - {20\choose0} - {20\choose1} - {20\choose2}$$

• oldguynewstudent

#### oldguynewstudent

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.