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In real situation, we identify a collection with certain characteristic common to elements. For example, a set of students in a class is based on the characteristic that each student is member of that class. This type of interpretation, however, is generally restrictive and leads to misinterpretation. We tend to think that the collection is isolated in itself, which is obviously wrong.
We need to free our mind from thinking set as an isolated entity. Some of the students might be members of another collection like that of basketball team, whereas some others might be members of a particular house, say “Amity house” and so on
In the nutshell, we consider set as a collection, which has multiple intersections with other collections.
Problem 1: In the house of total 200 students, 140 students play basketball and 80 students play football. Each student of the house plays at least one of these two games. How many students play both basketball and football?
Solution : The individual sets here are students playing basket ball (B) and football (F). Hence,
$$n\left(B\right)=140$$
$$n\left(F\right)=80$$
Clearly, there is no bar that a students playing basketball can not play football. This is also evident from the sum of the numbers in each set. The sum is 140 + 80 = 220, whereas total numbers of students in the house is 200 only. Thus, there are students who play both games. We can interpret the total numbers as the union of two individual sets. Hence, applying expansion for the numbers of a union :
$$n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)-n\left(A\cap B\right)$$
The students who play both games constitute the intersection of two individual sets.
Putting values,
$$\Rightarrow n\left(B\cup F\right)=140+80-200=20$$
Universal is inclusive of all related sets. If we observe the Venn’s diagram consisting of two individual sets, then we realize that largest closed region within the universal set is the union involving two sets i.e (A∪B). This union, however, is a subset of U. There is remaining area within the universal set, which is called the component of this union.
Now we know that a union represents elements which belong to either set exclusively or belong commonly with other sets. It means that the complement of union represents the region, which can not be defined by the characterizing criteria of the union. This complement of union, therefore, represents situations which is described in terms of “neither or nor” type. Actually, this set is given by De-morgan’s first law.
Problem 2: In a house of total 200 students, 100 students play basketball, 60 students play football and 20 play both games. How many students play neither basketball nor football?
Solution : We have already discussed that “neither nor” condition is same as that of De-morgan’s first law :
$$n\left(B\prime \cap F\prime \right)=n\left(B\cup F\right)\prime $$
Now expanding the right hand term, we have :
$$\Rightarrow n\left(B\prime \cap F\prime \right)=n\left(B\cap F\right)\prime =U-n\left(B\cup F\right)$$
Further using formula for the numbers in a union,
$$\Rightarrow n\left(B\prime \cap F\prime \right)=U-n\left(B\right)-n\left(F\right)+n\left(B\cap F\right)$$
Putting values,
$$\Rightarrow n\left(B\prime \cap F\prime \right)=200-100-60+20=60$$
This is the required answer. However, there remains a question : why do we consider total numbers of students as the numbers in universal set, “U”, unlike previous example in which this number corresponds to numbers in the union of individual sets. Remember, earlier question had the phrase “Each student of the house plays at least one of these two games”. This ensured that total numbers represented the union as everyone was playing one of two games. Such restriction is not there in this example. In fact, we saw that there are students who are not playing either of two games at all! Thus, total number represents universal set in this example.
Union of two sets “A” and “B” conveys the meaning of consisting three categories of elements (i) elements exclusively belonging to “A” (ii) elements exclusively belonging to “B” and (iii) (i) elements commonly belonging to “A” and “B”. In totality, we see that union conveys the meaning of “or” – the elements may belong either to a particular set or to both sets.
Problem 3: In a group of students, 40 students study either English or Mathematics. Of these 25 students study Mathematics, 10 students study both Mathematics and English. How many students study English?
Solution : The word “or” in the first sentence indicates that union of students studying Mathematics (M) or English (E) or both is 40. Using formula, we have :
$$n\left(M\cup E\right)=n\left(M\right)+n\left(E\right)-n\left(M\cap E\right)$$
$$\Rightarrow n\left(E\right)=n\left(M\cup E\right)-n\left(M\right)+n\left(M\cap E\right)$$
Putting values,
$$\Rightarrow n\left(E\right)=40-25+10=25$$
In the case of intersection of two sets, we have noted that difference represents the exclusive or isolated set, which is not common to other set. From the Venn’s diagram, we also observe that a given set is actually composed of two sets (i) difference set and (ii) intersection set.
$$n\left(A\right)=n\left(A-B\right)+n\left(A\cap B\right)$$
and
$$n\left(B\right)=n\left(B-A\right)+n\left(A\cap B\right)$$
Problem 4: In a house of 200 students, 120 students study Mathematics, 60 students study English and 40 students study both Mathematics and English. Find in the house : (i) students who study Mathematics but not English (ii) students who study English, but not Mathematics (iii) students who study either Mathematics or English and (iv) students who neither study Mathematics nor English.
Solution : Let us first characterize collections as given in the question. Two sets are given one for those who study Mathematics (M) and other for those who study English(E). The addition of numbers of individual sets is 120 + 60 = 180, which is less than total numbers of students. Hence, total numbers of 200 corresponds to universal set. Here,
$$U=200;\phantom{\rule{1em}{0ex}}n\left(M\right)=120;\phantom{\rule{1em}{0ex}}n\left(E\right)=60\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}n\left(M\cap E\right)=40.$$
(i) Students studying Mathematics, but not English means that we need to find the numbers in the difference of set i.e M – E.
$$n\left(M-E\right)=n\left(M\right)-n\left(M\cap E\right)$$
$$\Rightarrow n\left(M-E\right)=120-40=80$$
(ii) Students studying Mathematics, but not English means that we need to find the numbers in the difference of set i.e E – M.
$$n\left(E-M\right)=n\left(E\right)-n\left(M\cap E\right)$$
$$\Rightarrow n\left(E-M\right)=60-40=20$$
(iii) Students who study either Mathematics or English is equal to the numbers in the union of two sets.
$$n\left(M\cup E\right)=n\left(M\right)+n\left(E\right)-n\left(M\cap E\right)$$
Putting values,
$$\Rightarrow n\left(M\cup E\right)=120+60-40=140$$
(iv) Students who study neither Mathematics nor English is equal to the numbers in the compliment of the union of two sets.
$$n\left(M\cup E\right)\prime =U-n\left(M\cup E\right)=200-140=60$$
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