Three excursions

Each pupil of the 9A class attended at least one of the three excursions. There could always be 15 pupils on each excursion. Seven participants of the first excursion also participated in the second, 8 participants of the first excursion, and 5 participants of the second excursion also attended the third. Four pupils participated at all three excursions. How many pupils were in the 9A class?

Result

n =  29

Solution:

n1=15−(8+7−4)=4 n2=15−(7+5−4)=7 n3=15−(8+5−4)=6   n=n1+n2+n3+(7+8+5−3⋅ 4)+4=4+7+6+(7+8+5−3⋅ 4)+4=29n_{1}=15-(8+7-4)=4 \ \\ n_{2}=15 - (7+5-4)=7 \ \\ n_{3}=15 - (8+5-4)=6 \ \\ \ \\ \ \\ n=n_{1}+n_{2}+n_{3}+(7+8+5-3 \cdot \ 4) + 4=4+7+6+(7+8+5-3 \cdot \ 4) + 4=29

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