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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 2 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 2
Let the number of elements of the sets $$A$$ and $$B$$ be $$p$$ and $$q$$ respectively. Then, the number of relations from the set $$A$$ to the set $$B$$ is
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$${ 2 }^{ p+q }$$
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$${ 2 }^{ pq }$$
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$$p+q$$
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$$pq$$
Explanation
Given, number of elements of $$A$$ and $$B$$ are $$p$$ and $$q$$ respectively.
$$\therefore$$ The number of relations from the set $$A$$ to the set $$B$$ is $${2}^{pq}$$.
Find the second component of an ordered pair $$(2, -3)$$
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$$2$$
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$$3$$
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$$0$$
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$$-3$$
Explanation
In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in
an ordered pair $$(2,-3)$$, the second component is $$-3$$.
A ______ maps elements of one set to another set.
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order
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set
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relation
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function
Explanation
A relation map elements of one set to another set
$$R:A\to B$$
Elements in A is mapped to elements in set B.
The ______ product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set.
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cartesian
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coordinate
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simple
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discrete
Explanation
The cartesian product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set.
Example:$$ A = \{1, 2\} \quad B = \{2\}$$
cartesian product, $$A \times B = {(1,2), (2, 2)}$$
Identify the first component of an ordered pair $$(2, 1)$$.
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$$1$$
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$$2$$
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$$-1$$
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$$0$$
Explanation
In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in
an ordered pair $$(2,1)$$, the first component is $$2$$.
Cartesian product of sets $$A$$ and $$B$$ is denoted by _______.
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$$A \times B$$
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$$B \times A$$
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$$A \times A$$
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$$B \times B$$
Explanation
Cartesian product of Set $$A$$ and $$B$$ is denoted by $$A\times B$$.
What is the relation for the statement "A is taller than B"?
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is taller than
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A is taller
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B is taller
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is less than
Explanation
A is taller than B.
A is related to B by 'taller than'.
Identify the first component of an ordered pair $$(0, -1) $$.
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$$0$$
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$$-1$$
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$$2$$
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$$1$$
Explanation
In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in
an ordered pair $$(0,-1)$$, the first component is $$0$$.
If $$A = \{1, 2, 3\}, B = \{3, 4\}$$
find (A $$\times$$ B) $$\cup$$ (B $$\times$$ A)
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$$\{(1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (3, 4), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)\}$$
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$$\{(2, 3), (3, 3), (1, 4), (2, 4), (3, 4), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)\}$$
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$$\{(1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (3, 4), (3, 1), (4, 1), (4, 2), (4, 3)\}$$
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None of these
Explanation
$$ (A\times B) = \left \{ (1,3),(1,4),(2,3),(2,4),(3,3),(3,4) \right \} $$
$$ (B\times A) = \left \{ (3,1),(3,2),(3,3),(4,1),(4,2),(4,3) \right \} $$
$$ \Rightarrow (A\times B)\cup (B\times A)= \left \{ (1,3),(1,4),(2,3),(2,4),(3,3),(3,4),(3,1),(3,2),(4,1),(4,2),(4,3) \right \} $$
If A= {0, 1} and B ={1, 0}, then what is A x B equal to ?
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{(0, 1), (1, 0)}
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{(0, 0), (1, 1)}
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{(0, 1), (1, 0), (1, I)}
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A X A
Explanation
$$\left\{ { 0,1 } \right\} \times \left\{ 1,0 \right\} ={ \left\{ (0,1),(0,0),(1,1),(1,0) \right\} }$$
$$\left\{ { 0,1 } \right\} \times \left\{ 0,1 \right\} ={ \left\{ (0,0),(0,1),(1,0),(1,1) \right\} }$$
So, $$A\times B=A\times A$$
Hence, D is correct.
If $$A = \{a, b\}, B=\{1, 2, 3\}$$, find B $$\times$$ A
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$$B$$ $$\times$$ $$A$$$$ = \{(1, a), (2, a), (3, a), (1, b) (2, b), (3, b)\}$$
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$$B$$ $$\times$$ $$A$$$$ = \{ (2, a), (3, a), (1, b) (2, b), (3, b)\}$$
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$$B$$ $$\times$$ $$A$$$$ = \{(1, a), (2, a), (3, a), (1, b) (2, b)\}$$
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None of these
Explanation
To find B × A multiply each element of B with that of A & form an ordered pair.
i.e. ordered pairs are (1,a); (2,a); (3,a); (1,b); (2,b); (3,b)
Therefore B × A = {
(1,a), (2,a), (3,a), (1,b), (2,b), (3,b)}
Let R be the set of real numbers and the mapping $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ be defined by $$f(x)=5-x^2$$ and $$g(x)=3\lambda-4$$, then the value of $$(fog)(-1)$$ is
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$$-44$$
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$$-54$$
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$$-32$$
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$$-64$$
Explanation
Given, $$f(x)=5-x^2,\,g(x)=3x-4$$
Now, $$(fog)(-1)=f\begin{Bmatrix}g(-1)\end{Bmatrix}$$
$$=f(-7)\;[\because\;g(-1)=3(-1)-4=-7]$$
$$=5-(-7)^2$$
$$=-44$$
Let $$A = \left \{x, y, z\right \}$$ and $$B = \left \{p, q, r, s\right \}$$. What is the number of distinct relations from $$B$$ to $$A$$?
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$$4096$$
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$$4094$$
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$$128$$
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$$126$$
Explanation
Since, No. of elements in set A$$=3$$
No. of elements in set B$$=4$$
$$\therefore$$ No. of elements in B$$\times$$A$$=3\times 4=12$$
$$\therefore$$ No. of distinct relations from B to A $$= 2^{12}=4096$$
Option A is correct.
Let R be the relation in the set N given by = {(a, b): a = b - 2, b > 6}. Choose the correct answer
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$$(2, 4) \epsilon R$$
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$$(3, 8) \epsilon R$$
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$$(6, 8) \epsilon R$$
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$$(8, 7) \epsilon R$$
Explanation
Firstly b should be greater than 6 & also difference between b& a should be 2 , option C is satisfing both the conditions hence the correct answer is C
Let $$A=\left\{ u,v,w,z \right\} ;B=\left\{ 3,5 \right\} $$, then the number of relations from $$A$$ to $$B$$ is
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$$256$$
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$$1024$$
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$$512$$
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$$64$$
Explanation
Given
$$A=\left\{ u,v,w,z \right\} ;B=\left\{ 3,5 \right\} $$
Here, number of element in set $$A$$ is $$n(A)=4$$
And number of element in set $$B$$ is $$ n(B)=2$$
$$\therefore$$ Number of relations from $$A$$ to $$B$$
$$={ 2 }^{ n(A).n(B) }={ 2 }^{ 4\times 2 }={ 2 }^{ 8 }=256$$
Let A be a finite set containing n distinct elements. The number of relations that can be defined on A is.
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$$2^n$$
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$$n^2$$
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$$2n^2$$
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$$2n$$
On the set $$N$$ of all natural numbers define the relation $$R$$ by $$a R b$$ if and only if the G.C.D. of $$a$$ and $$b$$ is $$2$$. Then $$R$$ is:
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Reflexive, but not symmetric
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Symmetric only
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Reflexive and transitive
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Reflexive, symmetric and transitive
Explanation
$$aRa$$ only if GCD of a and a is 2 which is clearly not the case.
Hence not reflexive.
$$aRb \Rightarrow bRa$$ if GCD of a and b is 2 then GCD of b and a is also 2.
The relation is symmetric.
Transitiveness is not necessary,i.e, if and b have GCD 2 and b and c also have GCD 2 a and c need not have GCD 2.
The relation is symmetric only.
Consider two sets $$A=\{a, b, c\}, B=\{e, f\}$$. If maximum numbers of total relations from A to B; symmetric relation from A to A and from B to B are $$l, m, n$$ respectively, then the value of $$2l+m-n$$ is
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$$212$$
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$$184$$
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$$240$$
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$$64$$
Explanation
We have $$n(A)=3$$ and $$n(B)=2$$.
Now cartesian product of $$A$$ and $$B$$ can contain maximum $$3\times 2=6$$ elements.
Since relation is the possible subsets of cartesian product.
So maximum number of relations from $$A$$ to $$B$$ be $$(l)=2^{6}=64$$.
Now maximum possible symmetric relation from $$A$$ to $$A$$ be $$(m)=2^{\cfrac{3(3+1)}{2}}=2^6=64$$ and that of from $$B$$ to $$B$$ be $$(m)=2^{\cfrac{2(2+1)}{2}}=2^3=8$$. [Using formula for number of symmetric relation]
Now $$2l+m-n=128+64-8=184$$.
The number of reflexive relation in set A = {a, b, c} is equal to
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$$2^9$$
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$$2^4$$
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$$2^7$$
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$$2^6$$
Explanation
$$\therefore$$ Total number of relation in set A $$2 ^{3 \times 3} = 2^9$$
and maximum number of cartesian product = 9 out of which 3 ordered pair is necessary for reflexive.
So, for remaining 6 ordered pair
Number of ordered pair required.$$=^6 C _0\, +\, ^6C_1 \,+ \, ^6C_2 +......^6C_6$$ =$$ 2^6 $$
If the relation is defined on $$R-\left\{ 0 \right\} $$ by $$\left( x,y \right) \in S\Leftrightarrow xy>0$$, then $$S$$ is ________
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an equivalence relation
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symmetric only
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reflexive only
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transitive only
Explanation
$$s\Leftrightarrow xy$$
for (x,x) on relation
$$s\Leftrightarrow xy$$
and $$x^2>0$$
so, $$S$$ is symmetric
if $$xy>0$$
thus $$xy>0$$
so, $$S$$ is reflexive.
Now, if $$xy>0$$
and $$yz>0$$
so, $$xz>0$$
Thus, $$S$$ is transitive.
So, relation is equivilance.
If $$A=\left\{2,4\right\}$$ and $$B=\left\{3,4,5\right\},$$ then
$$\left( A\cap B \right) \times \left( A\cup B \right)$$ is
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$$\left\{(2,2),(3,4),(4,2),(5,4)\right\}$$
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$$\left\{(2,3),(4,3),(4,5)\right\}$$
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$$\left\{(2,4),(3,4),(4,4),(4,5)\right\}$$
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$$\left\{(4,2),(4,3),(4,4),(4,5)\right\}$$
Explanation
From given we get,
$$A\cap B=\{ 4\} ,A\cup B=\{ 2,3,4,5\}$$
$$\therefore \left( A\cap B \right) \times\left( A\cup B \right)$$$$=\left\{ (4,2),(4,3),(4,4),(4,5) \right\} $$
Ans. $$(d)$$.
Let $$A=\left\{1,2,3,4,5\right\}, B=\left\{2,3,6,7\right\}.$$ Then the number of elements in $$\left( A\times B \right) \cap \left( B\times A \right)$$ is
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$$18$$
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$$6$$
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$$4$$
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$$0$$
Explanation
Ans. $$(c)$$.
Common elements will be
$$\left\{ { (2,2),(3,3),(2,3),(3,2) } \right\} $$
Let A={ 1, 2, 3, 4} and R= {( 2, 2), (3, 3), (4, 4), (1, 2)} be a relation on A. Then R is
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Reflexive
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Symmetric
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Transitive
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None of these
Explanation
$$\neq$$ R as every element is not related to itself.
$$\neq$$ S as (1, 2) $$\in$$ R but (2, 1) $$\notin$$ R
T (1,2), $$\in$$ R, (2, 2) $$\in$$ R
$$\Rightarrow $$ (1, 2)o (2, 2) = (1, 2) $$\in$$ R
If A = (a, b, c, d), B= (p, q, r, s). then which of the following are relations from A to B? Give reasons for your answer:
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$$R_1$$ = {( a, p), (b, r), (c, s)}
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$$R_2$$ = { ( q, b), (c, s), (d, r)}
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$$R_3$$ = {(a, p), (a, q), (d, p), (c, r), (b, r)}
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$$R_4$$ = {(a, p), (a, q), (b, s), (s, b)}
Explanation
(A) .(C)
If R is a relation from A to B, then R $$\subset (A\times B)$$ which consists of all ordered pairs of the type. (x, y) such that $$x\in A, y \in B$$.
Clearly, $$R_1$$ and $$R_3$$ are such sets of $$A\times B$$. The elements (q, b), of $$R_2 \in (B\times A)$$and not $$A\times B $$ and the elements (q, b), and (s, b), of $$R_4$$ also not belong to $$A\times B$$ and as such $$R_2$$ and $$R_4$$ are not relations from A to B.
Given the relation R= {(1,2), (2,3) } on the set {1, 2, 3}, the minimum number of ordered pairs which when added to R make it an equivalence relation is
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5
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6
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7
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8
Explanation
For R add ( 1,1) ( 2, 2) (3, 3) - 3
For S add (2, 1) (3, 2)- 2
For T add (1, 3) ( 3, 1) - 2
Thus the minimum number of ordered pairs to be added is 7.
The union of two equivalence relations is:
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always an equivalence relation
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reflexive and symmetric but needn't be transitive
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reflexive and transitive but need not be symmetric
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None of these
Explanation
The union of two equivalence relation is not necessarily an equivalence relation.
Union of reflexive relation is reflexive,
Also, the union of symmetric relation is symmetric.
But the union of a transitive relation is not necessarily transitive.
Hence, the union of two equivalence relation is not equivalence.
Let R be a relation from a set A to a set B then
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$$R = A \cup B$$
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$$ R = A \cap B$$
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$$ R \subseteq A\times B$$
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$$ R\subseteq B\times A$$
Explanation
Let R be a relation from a set A to a set B then
$$R\subseteq A\times B$$
If $$n(A) = 2, n(B) = m$$ and the number of relation from $$A$$ to $$B$$ is $$64$$, then the value of $$m$$ is
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$$6$$
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$$3$$
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$$16$$
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$$8$$
Explanation
The number of relations between sets can be calculated using $$ 2^{mn}$$ where m and n represent
the number
of members in each set.
Given that $$ n(A) = 2 $$ and $$ n(B) = m $$
So,
$$ 2^{2m} = 64 $$
$$ 2^{2m} = 2^{6} $$
On comparing powers of 2 we get,
$$ 2m = 6 $$
$$ m = 3 $$
If $$A$$ and $$B$$ are two sets containing four and two elements, respectively. Then the number of subsets of the set $$A\times B$$ each having at least three elements is
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$$219$$
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$$256$$
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$$275$$
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$$510$$
Explanation
$$A=\left\{a,b,c,d\right\}$$
$$B=\left\{x,y\right\}$$
$$A\times B=2\times 4=8$$ Elements.
Total number of subsets of $$A\times B$$ having $$3$$ or more elements $$=2^8=256$$
$$\Rightarrow 256-(1$$ null set $$+$$ $$8$$ singleton set $$+$$ $$^8e_2$$having $$2$$ elements$$)$$
$$=256-1-8-28$$
$$=219$$
Find x and y, if $$(x+3,5)=(6,2x+y)$$.
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x=3, y=-1
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x=6, y=2
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x=2,y=3
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none of these
Explanation
By the definition of equality of ordered pairs
$$(x+3,5)=(6,2x+y)$$
$$x+3=6$$ and $$5=2x+y$$
$$x=3$$ and $$5=2x+y$$
$$x=3$$ and $$5=6+y$$
$$x=3$$ and $$y=-1$$
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