we have learnt union, intersection and difference of two sets. In this Section, we will go through some practical problems related to our daily life.The formulae derived in this Section will also be used in subsequent Chapter on Probability.

Let `A` and `B` be finite sets. If `A ∩ B = phi`, then

`(i)` `n ( A ∪ B ) = n ( A ) + n ( B )` ................ `(1)`

The elements in `A ∪ B` are either in `A` or in `B` but not in both as `A ∩ B = phi.` So, `(1)` follows immediately.

In general, if `A` and `B` are finite sets, then

`(ii)` `n ( A ∪ B ) = n ( A ) + n ( B ) - n ( A ∩ B )`.......................... `(2)`

Note that the sets `A - B, A ∩ B` and `B - A` are disjoint and their union is `A ∪ B` (Fig). Therefore

`n ( A ∪ B) = n ( A - B) + n ( A ∩ B ) + n ( B - A )`

`\ \ \ \ \ \ \ \ \ \ = n ( A - B) + n ( A ∩B ) + n ( B - A ) + n ( A ∩ B ) - n ( A ∩ B)`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= n ( A ) + n ( B ) - n ( A ∩ B),` which verifies `(2)`

`(iii)` If `A, B` and `C` are finite sets, then

`n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C ) - n ( A ∩ B ) - n ( B ∩ C) - n ( A ∩ C ) + n ( A ∩ B ∩ C )` ................. `(3)`

In fact, we have

`n ( A ∪ B ∪ C ) = n (A) + n ( B ∪ C ) - n [ A ∩ ( B ∪ C ) ]` [ by (2) ]

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = n (A) + n ( B ) + n ( C ) - n ( B ∩ C ) - n [ A ∩ ( B ∪ C ) ] ` [ by (2) ]

Since `A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ),` we get

`n [ A ∩ ( B ∪ C ) ] = n ( A ∩ B ) + n ( A ∩ C ) - n [ ( A ∩ B ) ∩ (A ∩ C)]`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \= n ( A ∩ B ) + n ( A ∩ C ) - n (A ∩ B ∩ C)`

Therefore

`n ( A ∪ B ∪ C ) = n (A) + n ( B ) + n ( C ) - n ( A ∩ B ) - n ( B ∩ C)- n ( A ∩ C ) + n ( A ∩ B ∩ C )`