Cardinality of a set
The cardinality of set A is defined as the number of elements in the set A and is denoted by n(A).
For example, if A = {a,b,c,d,e} then cardinality of set A i.e.n(A) = 5
Let A and B are two subsets of a universal set U. Their relation can be shown in Venndiagram as:
$$ n(A) = n_o( A) + n(A \cap B)$$
$$\text{or,}\: n(A)  n (A \cap B)= n_o(A)$$
$$ n(B) = n_o(B) + n(A \cap B)$$
$$\text {or,}\: n(B)  n(A \cap B) = n_o(B)$$
Also,
\begin{align*} n(A∪B) &= n_o(A) + n(A∩B) + n_o(B)\\ n(A∪B) &= n(A)  n(A∩B) + n(A∩B) + n(B)  n(A∩B)\\ n(A∪B) &= n(A) + n(B) n(A∩B)\\ \therefore n(A∪B) &= n(A) + n(B)  n(A∩B)\\ \end{align*}
If A and B are disjoint sets then:
\(n(A \cap B) = 0, n(A \cup B) =n(A) + n(B)\)
Again,
\(n(U) = n(A \cup B) + n(\overline {A\cup B)}\)
If \(n(\overline {A \cup B)}\)=0, then \( n(U) = n(A \cup B)\)
Problems involving three sets
Let A, B and C are three nonempty and intersecting sets, then:
\(n(A \cup B \cup C) = n(A) + n(B) +n(C)  n(A \cap B) n(B \cap C) n(C \cap A) +n(A \cap B \cap C).\)
In Venndiagram
\(n(A)\) = Number of elements in set A.
\(n(B)\) = Number of elements in set B.
\(n(C)\)=Number of element in set C.
\(n_o(A)\) = Number of elements in set A only.
\(n_o(B)\) = Number of elements in set B only.
\(n_o(C)\) = Number of elements in set C only.
\(n_o(A \cap B)\) = Number of elements in set A and B only.
\(n_o(B \cap C)\) = Number of elements in set B and C only.
\(n_o(C \cap A)\) = Number of elements in set A and C only.
\(n(A \cap B \cap C)\) = Number of elements in set A, B and C.
From the Venndiagram
\begin{align*} n(A \cup B \cup C) &= n_o(A) +n_o(B) +n_o(C) +n_o(A \cap B) +n_o(B \cap C) +n_o(C \cap A) + n(A \cap B \cap C)\\ &= n(A)  n_o(A \cap B)  n_o(C \cap A)  n(A \cap B \cap C) + n(B)  n_o(B \cap C)  n_o(C\cap B)  n(A \cap B \cap C)
+ n(C)  n_o(A \cap C)  n_o(B \cap C)  n(A \cap B \cap C) + n_o(A \cap B) +n_o(B \cap C) +n_o(C \cap A) + n(A \cap B \cap C)\\ &= n(A) + n(B) + n(C)  [n_o(A \cap B) +n(A \cap B \cap C)]  [n_o(A \cap B) +n(A \cap B \cap C)]  [n_o(B \cap C) +n(A \cap B \cap C)]  [n_o(C \cap A) +n(A \cap B \cap C)]+n(A \cap B \cap C)\\ &= n(A) + n(B) + n(C)  n(A \cap B) n(B \cap C) n(A \cap C) +n(A \cap B \cap C)\\ \end{align*}
$$\boxed{\therefore (A \cup B \cup C)= n(A) + n(B) + n(C)  n(A \cap B) n(B \cap C) n(C \cap A) +n(A \cap B \cap C)} $$
If A, B and C are disjoint sets,
\(n(A \cup B \cup C) = n(A) + n(B) + n(C)\)
 The cardinality of a set is a positive integer but it is not decimal. So, n(A) is not equal to 50% because 50% = 0.5.
 If A, B and C are disjoint sets, \(n(A \cup B \cup C) = n(A) + n(B) + n(C).\)
 \((A \cup B \cup C)= n(A) + n(B) + n(C)  n(A \cap B) n(B \cap C) n(C \cap A) +n(A \cap B \cap C)\)
Solution:
n(U) = 43
n(A) = 25
n(B) = 18
n(A ∩ B)= 7
The Venndiagram of above information is as follow:
\begin{align*}\overline {A \cup B} &= ? \\ n(A \cup B) &= n(A) + n(B)  n(A \cap B) \\ &= 25 + 18 7 \\ &= 43  7 \\ &= 36\\ Again, n(\overline {A \cup B}) &= n(U)  n(A \cup B) \\ &= 43  36\\ &= 7 Ans. \end{align*}
Solution:
Let N and H be the set of people who like to see Nepali movies and Hindi movies respectively
From question, n(U) = 80, n(N) = 47, n(H) = 31 , \((\overline {N \cup H})\) = 21
(i)
\begin{align*} n(N \cup H) &= n(U)  (\overline {N \cup H}) \\ &= 80  21 \\ &= 59 \\ Again, n(N \cup H) &= n(N) + n(H)  n(N \cap H)\\ or \: 59 &= 47 + 31 n(N \cap H) \\ or, \: 59 &= 78 n(N \cap H) \\ or, \:n(N \cap H) &= 78  59 \\ &= 19 \\ n_o(N) &= n(N) n(N \cap H) \\ &= 47 19 \\ &= 28 Ans. \end{align*}
(ii)
The no. of people who like to see Hindi movies only
\begin{align*} n_o (H) &= n(H) n(N \cap H) \\ &= 31  19 \\ &= 12 \end{align*}
(iii)
The above information is shown in the Venndiagram.
solution:
Let F and V represent the sets of people who like football and volleyball respectively and U be the universal sets.
From question,
\(n(F) = 40, n(F \cap V) = 10, \: n(U) = n(F \cup V) = 65 , \: n(V) = ? \)
(i)
We know that, \begin{align*} n(F \cup V) &= n(F) + n(V)  n(F \cap V) \\ or, \: 65 &= 40 + n(V)  10 \\ or, \: 65 &= 30 + n(V)\\ or, \: n(V) &= 65  30 \\ \therefore n(V) &= 35 Ans. \end{align*}
(ii)
\( n_o (F) = n(F)  n(F \cap V) \: = 40 10 \: = 30 Ans. \)
(iii)
\( n_o (V) = n(V) n(F \cap V) \: = 35  10 \: = 25 Ans. \)
Solution:
Let, F & M denote the set of people who like folk and modern song respectively. U be the universal set. From question,
n(U) = 100
n(F) = 65
n(M) = 55
n(F∩M) = 35
(i)
The above information represents in Venn diagram as follow:
From Venndiagram
\begin{align*} n(F \cup M) &= n(F) + n(M)  n(F \cap M)\\ &= 65 + 55  35 \\&= 85 \: Ans. \end{align*}
(ii)
\begin{align*} n(\overline {F \cup M} ) &= 100  85 \\ &= 15 Ans. \end{align*}
Solution:
Let, C and V represent the sets of student who like to play cricket and volleyball. Let, U be the universal set.
From question, \( n(C) = 20, \: n(V) = 15, \: n(U) = n(C \cup V) = 30, \: n(C \cap V) = ? \)
We know that,
\begin{align*} n(C \cup V) &= n(C) + n(V)  n(C \cap V) \\ or, \: 30 &= 20 + 15 n(C \cap V) \\or, \:n(C \cap V) &= 35  30\\ \therefore \: n(C \cap V) &= 5 \: Ans. \end{align*}
The above information is shown in the following venndiagram.
Solution:
Let, T and C represent the set of students who like to drink tea & coffee respectively.
Let, U be the universal set.
From question, \(n(U) = 120, \: n(T) = 88, \: n(C) = 26, \: n(\overline{T \cup C})= 17, \: Let \: n(T \cap C)= x\)
The above information is shown in Venndiagram.
We know that,
\begin{align*} n(T \cup C) &= n(U)  n(\overline{T \cup C})\\ n(T \cup C) &= 120  17 \\ &= 103 \\ Again, \: n(T \cup C) &= n(T) + n(C)  n(T \cap C)\\ or, \: 103 &= 88 + 26 n(T \cap C) \\ or, \: 103 &= 114  n(T \cap C)\\ \therefore \:n(T \cap C) &= 114  103 \\ &= 11 \: Ans. \end{align*}
Solution:
Let, T and C denote sets of students who like tea and coffee respectively. Let, U be the universal sets.
From question, \( n(U)= 130,\: n(T)=70, \: n(C)=40, n(\overline {T \cup C})= 30 \)
We know that,
\begin{align*} n(T \cup C ) &= n(U)  n(\overline{T \cup C})\\ n(T \cup C ) &= 130  30 \\ &= 100 \: Ans. \\ Again, \\n(T \cup C ) &= n(T) + n(C)  n(T \cap C)\\ or, \: 100 &= 40 + 70 n(T \cap C)\\ or, \: 100 &= 110 n(T \cap C) \\ \therefore n(T \cap C) &= 10 \: Ans. \\ \end{align*}
The Venndiagram showing the given information as follows.
Solution:
Let, A and B denote the set of students used the autorickshaw and bus. Let, U be the universal set,
From question,
\( n(U) = 131, \: n(A) = 56, \: n(B) = 103, \: n_o (B) = 65 \\ Let, n(A \cap B )= x, \: n(\overline{A \cup B}) = y \)
The above information is shown in Venndiagram as follows.
From Venndiagram,
\begin{align*} x + 65 &= 103 \\ or, \: x &= 103  65 \\ \therefore x= 38 \\ \end{align*}
(i) \begin{align*} \text {The no. of students used autorickshaw only} \: n_o (A) &= 56  x \\ &= 56  38 \\ &= 18 \: Ans. \end{align*}
(ii)\begin{align*} n(\overline {A \cup B}) &= n(U)  n(A \cup B)\\ &= 131  (65 + 38 + 18) \\ &= 131  121\\ &= 10 \: Ans. \end{align*}
Solution:
Let, B and L denote the set of tourist who like to visit Bhaktapur and Lalitpur respectively. Let, U be the universal set.
From question,
\( n(U) = 2400, \: n(B) = 1650, \: n(L) = 850, \: n(\overline{B \cup L}) = 150 \)
(i)
The above information is shown in Venndiagram as follows.
(ii)
\begin{align*} n(B \cup L) &= n(U)  n(\overline{B \cup L})\\ or, \:n(B \cup L) &= 2400  150 \\ &= 2250 \\ Again, \\n(B \cup L) &= n(B) + n(L)  n(B \cap L) \\ or, \: 2250 &= 1650 + 850 n(B \cap L)\\ or, \:n(B \cap L) &= 2500  2250 \\ \therefore \: n(B \cap L) &= 250 \: Ans. \\ \end{align*}
(iii)
\begin{align*} n_o (B) &= n(B)  n(B \cap L)\\ &= 850 250 \\ &= 600 \: Ans. \end{align*}
Solution:
Let, M and S denote the set of students who passed in mathematics and Science respectively. Let, U be the universal set,
From question,
\( n(M) = 60 \: n(S) = 45, \: n(M \cap S )= 30\)
(i) \begin{align*} n_o (M) &= n(M)  n(M \cap S)\\ &= 60 30 \\ &= 30 \: Ans. \end{align*}
(ii)\begin{align*} n_o (S) &= n(S)  n(M \cap S)\\ &= 45 30 \\ &= 15 \: Ans. \end{align*}
(iii) \begin{align*}n(M \cup S) &= n(M) + n(S) n(M \cap S)\\ &= 60 + 45  30 \\ &= 75 \: Ans. \end{align*}
(iv)The above information is shown in Venndiagram as follows.
Solution:
Let, M and E denote the set of students who like Maths and English respectively. Let, U be the universal set.
From question,
\( n(U)=55, \: n_o(M) = 15, \: n_o(E) = 18, \: n(\overline{M \cup E} )= 5 \\ Let \: n(M \cap E)= x \)
Now,
\begin{align*} n(M \cup E) &= n(U)  n(\overline{M \cup E})\\ &= 55 5 \\ &= 50 \end{align*}
The above information is shown in Venndiagram as follows.
From Venndiagram,
\begin{align*} 15 + x + 18 &= 50\\ or, \: x + 33 &= 50\\ or, \: x &= 50  33 \\ &= 17 \end{align*}
\( \therefore \text {The no. of students who like both subject is 17.} \)
Solution:
Let, M and S denote the set of students who like Mathematics and Science respectively. Let, U be the universal set,
From question,
\( n(U)=50 \: = n(M \cup S), \: n(M \cap S) = 20, \: Let, \: n(M)=3x \: and \: n(S)=2x \)
We know that,
\begin{align*} n(M \cup S) &= n(M) + n(S)  n(M \cap S)\\ 50 &= 3x + 2x  20 \\ or, \: 5x &= 50 + 20\\ or, \: x &= \frac{70}{5}\\ &= 14 \end{align*}
(i)\begin{align*} \text {The no. of student who like Mathematics, n(M)} &= 3x \\ &= 3 \times 14 \\ &= 42 \: Ans. \end{align*}
(ii)\begin{align*} \text {The no. of student who like Science, n(S)} &= 2x \\ &= 2 \times 14 \\ &= 28 \end{align*}
So, \( n_o(S) = n(S)  n(M \cap S) \: \:= 28  20 \: \: = 8\: Ans. \)
(iii) The above information is shown in Venndiagram as follows:
Solution:
Let, M and B denote the set of students who have taken Mathematics and Biology respectively. Let, U be the universal set.
From question,
\( n(M \cup B)=25 = n(U), \: n(M) = 12, \: n_o(M) = 8 \)
Now,
\begin{align*} n(M \cap B)&= n(M)  n_o(M) \\ &= 12  8 \\ &= 4 \end{align*}
\begin{align*} n_o(B) &= n(M \cup B)  n(M) \\ &= 25  12 \\ &= 13 \end{align*}
The no. of students who have taken Maths & Biology = 4 Ans.
The no. of students who have taken Biology but not Math = 13 Ans.
The above information is shown in Venndiagram as follows:
Solution:
Let, M and S denote the set of students who passed in mathematics and Science respectively. Let, U be the universal set,
From question,
\( n_o(M) = 40\%, \: n_o(S) = 30\%, \: n(\overline{M \cup S })= 10\%, \: n(U) = 100\% \)
\begin{align*}n(M \cup S)&= n(U) n(\overline{M \cup S }) \\ &= 100\%  10\% \\ &= 90\% \end{align*}
(i)
\begin{align*} n(M \cup S) &=n_o(M) +n_o(S) + n(M \cap S)\\ 90\% &= 40\% + 30\% + n(M \cap S)\\ or, \: 90\%  70\% &= n(M \cap S)\\ \therefore n(M \cap S) &= 20\% \end{align*}
(ii)
\begin{align*} n(M) &= n_o(M) + n(M \cap S)\\ &= 40\% + 20\% \\ &= 60\% \end{align*}
(iii)
The above information is shown in Venndiagram as follows.
Solution:
Let F & M denote the set of people who liked folk & modern song respectively.
Let U be the universal set. From question,
\( n(F)= 70\% , \: n(M)= 60\%, \: n(\overline{F \cup M})= 10\%, \: n(F \cap M)= 4000, \: n(U)= 100\% \)
The above information represents in Venndiagram as follows:
\begin{align*} n(F \cup M) &= n(U)  n(\overline{F \cup M}) \\ &= 100\% 10\% \\ &= 90\% \end{align*}
From Venndiagram
\begin{align*} 70\%  x\% + x\% + 60\%  x\% &=90\% \\ or, \: x &= (130  90)\% \\ \therefore x &= 40\% \end{align*}
Let total number of people be 'y'
\begin{align*} 40\% \; of\; y &= 4000 \\ or, \: y \times \frac{40}{100} &= 4000\\ or, \: y &= \frac{4000 \times 100}{40} \\ \therefore y &= 10000 \: Ans. \end{align*}
Solution:
From question, \( n(U)=30, \: n(A)=20,\: n(B)= 10 \)
\( 3x + y = n(A) \\ 3x + y = 20 \: \: \: \: \: .............(1)\)
\( x + y = n(B) \\ x + y = 10 \: \: \: \: \: \: ..............(2)\)
Subtracting eqn (2) from eqn (1)
\begin{array}{rrrr} 3x&+ &y&=&20 \\ x&+&y&=&10 \\ &&&&\\ \hline\\ &&2x&=&10\\\end{array}
\begin{align*} x &= \frac{10}{2} \\ \therefore x &= 5 \end{align*}
Putting value of x in eq^{n} (1)
\begin{align*} 3x + y &= 20 \\ or, \: 3 \times 5 + y &= 20 \\ or, \: y &= 20  15 \\ y &= 5 \end{align*}
(i) \begin{align*} n(A \cap B) &= y\\ &= 5 \: Ans.\end{align*}
(ii) \begin{align*} n(A \cup B) &= 3x + y + x \\ &= 3 \times 5 + 5 + 5 \\ &= 15 + 10\\ &= 25 \: Ans. \end{align*}
(iii) \begin{align*}n(\overline {A \cup B}) &= n(U)  n(A \cup B) \\ &= 30  25 \\ &= 5 \end{align*}
Solution:
Let, F and M be the sets of people who liked folk and modern songs.
Let U be the universal set.
From Question,
\( n(U) = 100\%, \: n(F) = 77\%, \: n(M) = 63\%, \: n(\overline {F \cup M} )= 5\% \), n(M∩N) = 135
\( Let, \: n(M \cap N) = x\% \)
We know that,
\begin{align*} n(F \cup M) &= n(U)  n(\overline {F \cup M})\\ &= 100\%  5\% \\ &= 95\% \\ Again,\\ n(F \cup M) &= n(F) + n(M)  n(F \cap M)\\ 95\% &= 77\% + 63\% x\%\\ or, \: x\% &= 140\%  95\%\\ \therefore x &= 45\%\end{align*}
(i) From question,
\begin{align*} 45\% \: of \: n(U) &= 135\\ or, \: n(U) \times \frac{45}{100} &= 135\\ or, \: n(U) &= \frac {135 \times 100}{45}\\ \therefore n(U) &= 300 \: Ans.\end{align*}
(ii)
\begin{align*}n(F) &= 300 \:of \: 77\%\\ &= 300 \times \frac {77}{100}\\ &= 231\\ Now, \\ n_o (F) &= n(F)  n(F \cap M)\\ &= 231  135 \\ &=96 \: Ans. \end{align*}
(iii)
The above information represent in Venndiagram as follow:
Solution:
Let M and S be the student who liked math and science respectively.
Let U be the universal sets then,
From question,
\( n(U)= 95, \: Let\: n(M)= 4x, \: n(S)= 5x, \: n(M \cap S)= 10, \: n(\overline{M \cup S})= 15 \)
The above information is shown in the venndiagram.
\begin{align*} n(M \cup S) &=n(U)  n(\overline{M \cup S})\\ &= 95 15 \\ &= 80 \end{align*}
\begin{align*} n(M \cup S) &=n_o(M)+n_o(S) + n(M \cap S)\\ 80 &= 4x  10 + 10 + 5x 10\\ or, \: 80 &= 9x  10\\ \therefore x &= \frac{90}{9} &= 10 \end{align*}
(a) The number of students who liked mathematics only
\begin{align*} 4x  10 &= 4 \times 10  10 \\ &= 40  10 \\ &= 30 \: Ans. \end{align*}
(b) The number of students who liked science only
\begin{align*} 5x  10 &= 5 \times 10  10 \\ &= 50  10 \\ &= 40 \: Ans. \end{align*}
Solution:
Let E, M and N represents the set of students who passed in English, Mathematics and Nepali respectively. Let U be the universal set.
\( n(U) = 200, \: \: \: \: \: n(E) = 70, \: \: \:\: \: n(M) = 80\: \:\: \: \: n(N) = 60, \\ n(E \cap M)=35, \: \:\: \: \: n(E \cap N)=25 \: \:\: \: n(M \cap N)=35 \: \:\: \: n(E \cap M \cap N) = 10 \)
(a) The above information is shown in venndiagram as follows.
(b) \begin{align*} n(E \cap M \cap N) &= n(E) + n(M) + n(N)  n(E \cap M)  n(M \cap N)  n(N \cap E) + n(E \cap M \cap N)\\ &= 70 + 80 + 60  35 25 35 + 10 \\ &= 125 \end{align*}
\begin{align*} n(\overline{E \cap M \cap N}) &= n(U) n(E \cap M \cap N) \\ &=200  125 \\ &= 75 \: Ans. \end{align*}
Let E, A and S denote the set of students who failedin English, Account and statistics respectively. Let U be the universal set.
From question,
\( n(U) = 100\% , \: n(E)=58\%, \: n(A) = 39\%, \: n(S)= 25\%, \: n(E \cap A)=32\%, \\ \: n(A \cap S)= 17\%, \: n(E \cap S)=19\% \: n(E \cap A \cap S)= 13\% \)
(a)The above information is shown in the venndiagram as follow:
\begin{align*}n(E \cup A \cup S) &= n(E) + n(A) + n(S)  n(E \cap A)  n(A \cap S)  n(S \cap E) + n(E \cap A \cap S)\\ &= (58 + 39 + 25  32  17  19  13)\% \\ &= 67\% \end{align*}
(b) \begin{align*} n(\overline {E \cup A \cup S}) &= n(U)  n(E \cup A \cup S )\\ &= 100\% 67\% \\ &= 33\% \: Ans. \end{align*}

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Jan 22, 2017 
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Shajan panditIn a group of students 20 study account, 21 study maths, 18 study history, 7 study account only, 10 study maths only, 6 study account and maths only, 3 study maths and history only. i) How many students study all subjects?ii) How many students are there altogether? 
Dec 31, 2016 
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