n(A) n(U)
ab n(B)
c
d
Derivation of formula n( ∪ ) = n( ) + n( )- n( ∩ )
Derivation
From above venn diagram, we have By Ram Krishna Marahatta
n( ∪ ) = a + b + c
n( ) = a + b
n( ) = b + c
adding (ii) and (iii) we get ,
n( ) + n( ) = a + b + b + c
n( ) + n( ) = ( a + b + c ) + b
n( ) + n( ) = n( ∪ ) + n( ∩ )
n( ∪ )= n( ) + n( ) - n( ∩ ) derived
n(A) n(U)
a n(B)
db
fge
c h
N(C)
Derivation of n( ∪ ∪ ) = n(A)+ n(B)+ n(C) − n( ∩ ) − ( ∩ ) − n( ∩ )+ n( ∩ ∩ )
Derivation
We have By Ram Krishna Marahatta
n( ∪ ∪ ) = a + b + c + d + e + f + g
n( ∩ ) = d + g
n( ∩ ) = e + g
n( ∩ ) = f + g
n(A) = a + d + f + g
n(B) = b + d + e + g
n(C)= c + e + f + g
n( ∩ ∩ ) = g
adding last four we get
n(A)+ n(B)+ n(C)+ n( ∩ ∩ ) = a + b + c + 2d +2 e + 2f + 3g
n(A)+ n(B)+ n(C)+ n( ∩ ∩ ) = (a + b + c + d +2 e + f + g) + ( d + g) + (e + g)
n(A)+ n(B)+ n(C)+ n( ∩ ∩ ) = n( ∪ ∪ ) + n( ∩ ) + n( ∩ ) + n( ∩ )
n( ∪ ∪ ) = n(A)+ n(B)+ n(C) − n( ∩ ) − ( ∩ ) − n( ∩ )+ n( ∩ ∩ ) derived
Derivation of ( )= n(A)− n( ∩ ) − n( ∩ ) + n( ∩ ∩ )
Derivation
From above venn-diagram We have By Ram Krishna Marahatta
n( ∩ ) = d + g
n( ∩ ) = f + g
n(A) = a + d + f + g
n( ∩ ∩ ) = g
adding we get
n(A) + n( ∩ ∩ ) = a + d + f + 2g
n(A) + n( ∩ ∩ ) = a + ( d + g ) + (f + g)
n(A) + n( ∩ ∩ ) = ( ) + n( ∩ ) + n( ∩ )
( )= n(A)− n( ∩ ) − n( ∩ ) + n( ∩ ∩ ) derived
Derivation
n( ∩ ) = d + g
n( ∩ ) = f + g
n(A) = a + d + f + g
n( ∩ ∩ ) = g
adding we get
n(A)+ n( ∩ ∩ ) = a + d + f + 2g
n(A)+ n( ∩ ∩ ) = a + ( d + g ) + (f + g)
n(A)+ n( ∩ ∩ ) = ( ) + n( ∩ ) + n( ∩ )
( )= n(A)− n( ∩ ) − n( ∩ ) + n( ∩ ∩ ) proved