Tag: cartesian product of sets

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Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

If A and B are two non-empty sets having n elements in common, then what is the number of common elements in the sets $A\times B$ and $B\times A$?

  1. $n$
  2. $n^2$
  3. $2n$

  4. Zero
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Say A has x elements and B has y elements in total. 


Their cartesian product $A\times B$ will have $x\times y$ elements.

Hence if they have n elements in common. $n^2$ common elements are present in the products $A\times B$ and $B\times A$

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

Let $A=\left{ x\in W,the\quad set\quad of\quad whole\quad numbers\quad and\quad x<3 \right} $

$B=\left{ x\in N,the\quad set\quad of\quad natural\quad numbers\quad and\quad 2\le x<4 \right} $ and $C=\left{ 3,4 \right} $, then how many elements will $\left( A\cup B \right) \times C$ conatin?

  1. $6$
  2. $8$
  3. $10$
  4. $12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$A={0,1,2}, B={2,3}$ and $C={3,4}$
$A \cup B={0,1,2,3}$
No. of elements in $(A\cup B)\times C$ $=$ No. of elements in $(A \cup B)\times $ No. of elements in $C$$=4\times 2=8$

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

Let $A = \left{ a,b,c,d \right}$ and $ B=\left{ x,y,z \right}$. What is the number of elements in $ A\times B$?

  1. $6$
  2. $7$
  3. $12$
  4. $64$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given sets are $A={a,b,c,d}$ and ${x,y,z}$ 

So $A\times B={(a,x),(a,y),(a,z),(b,x),(b,y),(b,z),(c,x),(c,y),(c,z),(d,x),(d,y),(d,z)}$
No. of elements in $A\times B$ is $3\times 4=12$

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

If $A = \left{ 1,2 \right}$, $B = \left{ 2,3 \right}$ and $ C = \left{ 3,4 \right}$, then what is the cardinality of $ \left( A\times B \right) \cap \left( A\times C \right) $

  1. $8$
  2. $6$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$A = \left\{ 1,2 \right\}$, $B = \left\{ 2,3 \right\}$ and $ C = \left\{ 3,4 \right\}$

Now, $A\times B=\{ (1,2),(1,3),(2,2),(2,3)\}$ 

$A\times C=\{ (1,3),(1,4),(2,3),(2,4)\} $ And

$ (A\times B)\cap (A\times C)=\{ (1,3),(2,3)\} $

So cardinality is $2$.

Hence, option C is correct.

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

If two sets $A$ and $B$ are having $39$ elements in common, then the number of elements common to each of the sets $A\times B$ and $B\times A$ are

  1. ${ 2 }^{ 39 }$
  2. ${ 39 }^{ 2 }$
  3. $78$
  4. $351$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If set $A$ and set $B$ have $39$ common elements, then the number of common elements in set $A\times B$ and set $B\times A\,=39^2$

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

If ${ y }^{ 2 }={ x }^{ 2 }-x+1$ and $\quad { I } _{ n }=\int { \cfrac { { x }^{ n } }{ y }  } dx$ and $A{ I } _{ 3 }+B{ I } _{ 2 }+C{ I } _{ 1 }={ x }^{ 2 }y$ then ordered triplet $A,B,C$ is

  1. $\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right) $
  2. $\left( 3,1,0 \right) $
  3. $\left( 1,-1,2 \right) $
  4. $\left( 3,-\cfrac { 5 }{ 2 } ,2 \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have given


${y^2} = {x^2} - x + 1$

${I _n} =\displaystyle \int {\dfrac{{{x^n}}}{y}dx} $
And,

$A{I _3} + B{I _2} + C{I _1} = {x^2}y$

Order triplet $A,\, B,\,C$ $ = \left( {\dfrac{1}{2},\,\dfrac{{ - 1}}{2},1} \right)$

Hence, the option $(A)$ is correct.

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product
Suppose $S=\{1,2\}$  and $T=\{a,b\}$  then $T \times S$
  1. <span class="math">$(a,1),(a,2),(b,1),(b,2)$
    <span class="MJX_Assistive_MathML">B×A={(a,1),(a,2),(b,1),(b,2)}
  2. $(1,a),(2,b),(b,1),(b,2)$
  3. $(a,1),(a,2),(1,b),(2,b)$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given : $S=\{ 1,2\} ,T=\{ a,b\} $
$T\times S$ is the set of ordered pair of elements of $T$ and $S$ respectively
Then, $T\times S=\{a,b\}\times \{1,2\}$
$T\times S=\{ (a,1),(a,2),(b,1),(b,2)\} $