Let \(A=\{1,2,3,4,5,6\}\) and \(B=\{0,1\}\). How many functions \(f:A\to B\) have at least \(3\) inputs whose image is \(1\)?

Correct answer: \(42\)

A function \(f:A\to B\) assigns each element of \(A\) either to image \(1\) or to image \(0\). We want at least 3 elements mapping to \(1\). If exactly \(k\) elements map to \(1\), then the number of choices is \(\binom{6}{k}\). So the total is
\[\binom{6}{3}+\binom{6}{4}+\binom{6}{5}+\binom{6}{6}
=20+15+6+1=42.\]