A∩B∩C´=c=7
A∩B=a+c=11⇒a=11−c=11−7=4
A’∩B∩C’=f=4
A∩B’∩C’=g=5 (this condition was assumed in order to solve a question)
B=a+b+c+f=18⇒b=18−a−c−f=18−4−7−4=3
A=a+c+g+d=16⇒d=16−a−c−g=16−4−7−5=0
C=a+b+d+e=17⇒e=17−a−b−d=17−4−3−0=10
A∪B∪C=a+b+c+d+e+f+g=4+3+7+0+10+4+5=33
x=33,x’=46−33=13 students choose no subject
1) Only one of the subject preferred: y=e+f+g=10+4+5=19 students
2) At most one subject preferred: y+x’=19+13=32 students
3) At least two subjects preferred: a+b+c+d=7+3+4+0=14 students
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