We can achieve this result analytically as well. Here, we consider “A” as one set and “ B ∪ C B ∪ C “ as other set. Then, we apply the relation, which has been obtained for the numbers in the union of two sets as :

n A ∪ B ∪ C = n A + n B ∪ C − n [ A ∩ B ∪ C ] n A ∪ B ∪ C = n A + n B ∪ C − n [ A ∩ B ∪ C ]

Applying result for the union of two sets for “ n B ∪ C n B ∪ C ”, we have :

n B ∪ C = n B + n C − n B ∩ C n B ∪ C = n B + n C − n B ∩ C

Putting in the expression for “ n A ∪ B ∪ C n A ∪ B ∪ C ”,

⇒ n A ∪ B ∪ C = n A + n B + n C − n B ∩ C − n [ A ∩ B ∪ C ] ⇒ n A ∪ B ∪ C = n A + n B + n C − n B ∩ C − n [ A ∩ B ∪ C ]

At this stage, our task is to evaluate “ n [ A ∩ B ∪ C ] n [ A ∩ B ∪ C ] ”, Recall that we have worked with the distributive property of “intersection operator over union operator”. Following distributive property,

⇒ n [ A ∩ B ∪ C ] = n [ A ∩ B ∪ A ∩ C ] ⇒ n [ A ∩ B ∪ C ] = n [ A ∩ B ∪ A ∩ C ]

We can treat each of the terms in the small bracket on the right hand side of the above equation as a set. Applying relation obtained for the numbers in the union of two sets again, we have :

⇒ n [ A ∩ B ∪ C ] = n A ∪ B + n A ∪ C − n [ A ∩ B ∩ A ∩ C ] ⇒ n [ A ∩ B ∪ C ] = n A ∪ B + n A ∪ C − n [ A ∩ B ∩ A ∩ C ]

The last term in above equation is :

[ A ∩ B ∩ A ∩ C ] = A ∩ B ∩ C [ A ∩ B ∩ A ∩ C ] = A ∩ B ∩ C

Hence,

⇒ n [ A ∩ B ∪ C ] = n A ∪ B + n A ∪ C − n [ A ∩ B ∩ C ] ⇒ n [ A ∩ B ∪ C ] = n A ∪ B + n A ∪ C − n [ A ∩ B ∩ C ]

Now, putting this expression in the expression of the numbers in the union involving three sets and rearranging terms, we have :

⇒ n A ∪ B ∪ C = n A + n B + n C - n A ∩ B - n A ∩ C - n B ∩ C + n A ∩ B ∩ C ⇒ n A ∪ B ∪ C = n A + n B + n C - n A ∩ B - n A ∩ C - n B ∩ C + n A ∩ B ∩ C

In the nutshell, we find that numbers of elements in the union, here, is equal to the sum of numbers in the individual sets, minus elements common to two sets taken at a time, plus elements common to all three sets.

We observe that sum of the individual sets is less than total numbers of people , reading newspaper, in the town. Hence, total reading population represents universal set, “U”. The representations of these sets are shown on Venn’s diagram. Note that we have split the elements common to a pair of two sets in two parts (a) elements exclusive to intersection of two sets and (b) elements common to all three sets.

Figure 3: The regions common to sets.

Union of three sets

From the diagram,

(i) The required set is the region of “A” not common to “B” and “C”. This region represents elements, which are exclusive to set “A”. Thus, numbers of people reading only “A” is :

n 1 = 40000 − 3000 + 2000 + 2000 = 33000 n 1 = 40000 − 3000 + 2000 + 2000 = 33000

(ii) The required set is the region of “B” not common to “A” and “C”. This region represents elements, which are exclusive to set “B”. Thus, numbers of people reading only “B” is :

n 2 = 30000 − 3000 + 1000 + 2000 = 24000 n 2 = 30000 − 3000 + 1000 + 2000 = 24000

(iii) The required set is the remaining region of universal set “U” i.e. complement of the union of three sets. Now, proceeding as before, people who read newspaper “C” only is (see Venn’s diagram above):

n 3 = 100000 − 2000 + 1000 + 2000 = 5000 n 3 = 100000 − 2000 + 1000 + 2000 = 5000

Hence, the required number is :

Figure 4: The regions representing people, who read neither of three newspapers.

Union of three sets

n 4 = 100000 − 33000 + 24000 + 5000 + 3000 + 2000 + 1000 + 2000 n 4 = 100000 − 33000 + 24000 + 5000 + 3000 + 2000 + 1000 + 2000

⇒ n 4 = 30000 ⇒ n 4 = 30000

(i) Here we are required to find the numbers of people reading only “A”. It is clear that this set is part of the people, who do not read newspapers “B” and “C”. As discussed in the case of two sets, the numbers of people who read neither “B” and “C” is given by De-morgan’s first equation,

B ′ ∩ C ′ = B ∪ C ′ B ′ ∩ C ′ = B ∪ C ′

Figure 5: The region representing people, who read neither of two newspapers.

Intersection of two complement sets

For clarity, we have shown this region in the Venn’s diagram. We should realize that this intersection of two sets also includes people who read newspaper “A”. However, we are required to know numbers of people, who read newspaper “A” only. The exclusion people reading other newspaper as well are not part of our required set.

The remaining set (refer Venn’s diagram) represent area consisting of people who exclusively read newspaper “A” or who do not read any of three newspapers. Now, we need the intersection of set A with the remaining region to obtain the numbers, who read only newspaper “A”. Hence, required number is :

n 1 = n A ∩ B ′ ∩ C ′ n 1 = n A ∩ B ′ ∩ C ′

Using De-morgan's equation,

n 1 = n A ∩ B ′ ∩ C ′ = n [ A ∩ B ∪ C ′ ] n 1 = n A ∩ B ′ ∩ C ′ = n [ A ∩ B ∪ C ′ ]

⇒ n 1 = n [ A ∩ [ U − B ∪ C ] ] = n [ A ∩ U − A ∩ B ∪ C ] ] ⇒ n 1 = n [ A ∩ [ U − B ∪ C ] ] = n [ A ∩ U − A ∩ B ∪ C ] ]

⇒ n 1 = n [ A − A ∩ B ∪ C ] = n A − n [ A ∩ B ∪ C ] ⇒ n 1 = n [ A − A ∩ B ∪ C ] = n A − n [ A ∩ B ∪ C ]

Using distributive property of intersection over union,

⇒ n 1 = n A − n [ A ∩ B ∪ A ∩ C ] ⇒ n 1 = n A − n [ A ∩ B ∪ A ∩ C ]

Using formula of expansion of union of two sets,

⇒ n 1 = n A − [ n A ∩ B + n A ∩ C − n [ A ∩ B ∩ A ∩ C ] ] ⇒ n 1 = n A − [ n A ∩ B + n A ∩ C − n [ A ∩ B ∩ A ∩ C ] ]

⇒ n 1 = n A − [ n A ∩ B + n A ∩ C − n A ∩ B ∩ C ] ⇒ n 1 = n A − [ n A ∩ B + n A ∩ C − n A ∩ B ∩ C ]

We see that values of each term on the right hand side are given. Putting these values,

⇒ n 1 = 40000 − [ 5000 + 4000 − 2000 ] = 33000 ⇒ n 1 = 40000 − [ 5000 + 4000 − 2000 ] = 33000

(ii) Here we are required to find the numbers of people reading only “B”. Proceeding as before,

n 2 = n B − [ n B ∩ A + n B ∩ C − n B ∩ A ∩ C ] n 2 = n B − [ n B ∩ A + n B ∩ C − n B ∩ A ∩ C ]

⇒ n 2 = n B − [ n A ∩ B + n B ∩ C − n A ∩ B ∩ C ] ⇒ n 2 = n B − [ n A ∩ B + n B ∩ C − n A ∩ B ∩ C ]

Putting values,

⇒ n 2 = 30000 − [ 5000 + 3000 − 2000 ] = 24000 ⇒ n 2 = 30000 − [ 5000 + 3000 − 2000 ] = 24000

(iii) In order to find the numbers of people, who read neither of three newspapers, we first find union of three sets. The union represents people who read either of these newspapers – one, two or all three. Clearly, people, who do not read either of these papers constitute a complement of this union.

n 3 = A ∪ B ∪ C ′ = U − A ∪ B ∪ C n 3 = A ∪ B ∪ C ′ = U − A ∪ B ∪ C

Using expansion for the numbers in the union of three sets,

⇒ n 3 = U − [ n A + n B + n C − n A ∩ B − n A ∩ C − n B ∩ C + n A ∩ B ∩ C ] ⇒ n 3 = U − [ n A + n B + n C − n A ∩ B − n A ∩ C − n B ∩ C + n A ∩ B ∩ C ]

Putting values, we have :

⇒ n 3 = 100000 − [ 40000 + 30000 + 10000 − 5000 − 4000 − 3000 + 2000 ] = 30000 ⇒ n 3 = 100000 − [ 40000 + 30000 + 10000 − 5000 − 4000 − 3000 + 2000 ] = 30000