Cardinality of a set

Intersection
Intersection

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 Venn-diagram 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 non-empty and intersecting sets, then:
\(n(A \cup B \cup C) = n(A) + n(B) +n(C) - n(A \cap B) -n(B \cap C) -n(C \cap A) +n(A \cap B \cap C).\)

Intersection

In Venn-diagram

\(n(A)\) = 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 Venn-diagram

\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 Venn-diagram of above information is as follow:

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\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*}

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(iii)
The above information is shown in the Venn-diagram.

solution:

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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

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(i)
The above information represents in Venn diagram as follow:
From Venn-diagram
\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*}

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The above information is shown in the following venn-diagram.

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 Venn-diagram.

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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 Venn-diagram showing the given information as follows.

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Solution:

Let, A and B denote the set of students used the autorickshaw and bus. Let, U be the universal set,
From question,
\( n(U) = 131, \: n(A) = 56, \: n(B) = 103, \: n_o (B) = 65 \\ Let, n(A \cap B )= x, \: n(\overline{A \cup B}) = y \)

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The above information is shown in Venn-diagram as follows.

From Venn-diagram,
\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 Venn-diagram as follows.

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(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 Venn-diagram as follows.

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Solution:

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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 Venn-diagram as follows.

From Venn-diagram,
\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 Venn-diagram as follows:

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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 Venn-diagram as follows:

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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 Venn-diagram as follows.

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Solution:

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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 Venn-diagram as follows:

\begin{align*} n(F \cup M) &= n(U) - n(\overline{F \cup M}) \\ &= 100\% -10\% \\ &= 90\% \end{align*}

From Venn-diagram
\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 eqn (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 Venn-diagram as follow:

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Solution:

Let M and S be the student who liked math and science respectively.
Let U be the universal sets then,
From question,
\( n(U)= 95, \: Let\: n(M)= 4x, \: n(S)= 5x, \: n(M \cap S)= 10, \: n(\overline{M \cup S})= 15 \)

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The above information is shown in the venn-diagram.

\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 venn-diagram as follows.

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(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 venn-diagram as follow:

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\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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Shajan pandit

In 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?