Question

In a class of 45 pupils, 21 play chess, 23 play cards and 5 play both the games. Find
(i)How many do not play any of the games;
(ii)How many play chess only;

Answer 1

Hint: Here, we have to find the number of students do not play any of the games and number of students play chess only. We have to use the concept of sets. The collection of well-defined distinct objects is known as a set.

Formula used:
We will use the following formulas:
1.Cardinal number of union of sets \[n(A \cup B) = n(A) + n(B) - n(A \cap B)\] , where \[n(A)\] is the cardinal number of the set A, \[n(B)\] is the cardinal number of the set B, \[n(A \cap B)\] is the cardinal number of intersection of sets.
2.Cardinal number of Complementary set \[n{(A \cup B)^{'}} = n(U) - n(AUB)\], where \[n(U)\] is the cardinal number of the universal set.
3.Difference of two sets :\[n\left( {A-B} \right){\rm{ }} = n\left( A \right){\rm{ }}-n(A \cap B)\]

Complete step-by-step answer:
Let A denote the number of pupils who play chess, B denote the number of pupils who play cards.
We are given that the total number of students in a class \[n(U) = 45\].
Number of students who play chess only \[n(A) = 21\]
Number of students who play cards only \[n(B) = 23\]
Number of students who play both the games \[n(A \cap B) = 5\]
Cardinal number of union of sets is given by \[n(A \cup B) = n(A) + n(B) - n(A \cap B)\]
Substituting the values of \[n(A),n(B),n(A \cap B)\], we have \[n(A \cup B)\]
\[ \Rightarrow n(A \cup B) = 21 + 23 - 5\]
\[ \Rightarrow n(A \cup B) = 39\]
Cardinal number of Complementary set \[n{(A \cup B)^{'}} = n(U) - n(AUB)\]
Substituting the values of \[n(U),n(A \cup B)\], we have \[n{(A \cup B)^{'}}\]
\[ \Rightarrow n{(A \cup B)^{'}} = 45 - 39\]
\[ \Rightarrow n{(A \cup B)^{'}} = 6\]

Difference of two sets: \[n\left( {A-B} \right){\rm{ }} = n\left( A \right){\rm{ }}-n(A \cap B)\]
Substituting the values of \[n(A),n(A \cap B)\], we have \[n(A - B)\]
\[ \Rightarrow n(A - B) = 21 - 5\]
\[ \Rightarrow n(A - B) = 16\]
Therefore, the number of students who do not play any of the games is \[6\]and the number of students who play chess only is \[16\].


Note:The problem can also be solved using the representation of a venn diagram. A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Circles that overlap have a commonality while circles that do not overlap do not share those traits. Venn diagrams help to visually represent the similarities and differences between two concepts.