Choose at least one of five funfair rides

Choose an unordered subset of the five different rides. Each ride is either selected once or not selected. The source is not counting ordered visits or arbitrary repetitions.

For rides 1,2,3,4,5 respectively, the choices are take or do not take: 2 choices each. Therefore $(2\times2\times2\times2\times2)-1=2^5-1=31$, subtracting the single case of taking no rides. The source caption beneath the fifth factor repeats “4th ride”; that factor belongs to the fifth ride.

Taking 1,2,3,4,5 rides respectively gives $\binom51=5$, $\binom52=10$, $\binom53=10$, $\binom54=5$, $\binom55=1$. These cases are disjoint and exhaust all nonempty subsets, so the total is $5+10+10+5+1=31$.