Please note that my solution differs from the authors’ one (though in a trivial way only) given at the end

A company is considering 6 possible new computer systems and its systems manager would like to try out at most 3 of them. In how many ways can the systems manager choose the systems to be tried out?

\[\binom{6}{1} + \binom{6}{2} + \binom{6}{3} = \frac{6}{1} + \frac{6 \cdot 5}{2 \cdot 1} + \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 6 + 15 + 20 = 41\]

Authors’ answer: \[\binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3} = \frac{1}{1} + \frac{6}{1} + \frac{6 \cdot 5}{2 \cdot 1} + \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 1 + 6 + 15 + 20 = 42\]¹