​​n(A ∩ B ∩ C​)=6​, ​​n(A′ ∩ B′ ∩ C′​)=19​, so h=6, k=19



n(A ∩ C​)=13​ = g+h, and since h=6, g=13-6=7



n(B ∩ C​)=11​ = h+i, and since h=6, i=11-6=5 





​n(A ∩ B′​)=8​ = d+g, and since g=7, d=8-7=1



​​n(A)=15​ = d+e+g+h, and since d=1, g=7, and h=6, e=15-1-7-6 = 1   ​ ​

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n(B ∩ C′​)=3​ = e+f, and since e=1, f=3-1=2

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​n(B ∪ C​)=33 = e+f+g+h+i+j, and since e=1, f=2, g=7, h=6, i=5,
j = 33-1-2-7-6-5 = 12

​​

Edwin