Let A = set of students, read Maths B = set of students, read Physics C = set of students, read Chemistry
Now, `n(A)=21,n(B)=26,n(C)=29`
`n(AcapB)=14,n(BcapC)=15,n(CcapA)=12`
`n(A capBcapC)=8`
`:. n(AcupBcupC)=n(A)+n(B)+n(C)-n(AcapB)-n(BcapC)-n(C capA)+n(A capBcapC)`
`= 21 + 26 + 29 - 14 -15 - 12 + 8 = 43`.