Question

please do it I m waiting

Answer 1

Let nA represent no. of students passed in physics
nB for chemistry, nC for Maths
nA =24, nB=43,nC=37
n(A U B U C) = 50

n(A n B) =20

n(B n C)=29
n(A n C) =19

by formula
n(A U B U C) = nA +nB+ nC -n(A nB) -n(B nC) - n(A n C) +n(A n B n C)
therefore n(A n B n C) = 20 + 29 +19 +50 +24 + 43 + 37=14
therefore 14 students passed in all 3 subjects.