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CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Set Theory
Quiz 12
State true or false for each of the following. Correct the wrong statement
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0%
True
0%
False
Explanation
Given
$$n (φ) = 1 $$
The statement given here is false
Correct statement: $$n (φ) = 0$$
In $$n(P)= n(M)$$, then $$P \rightarrow M$$
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0%
True
0%
False
Explanation
$$n(P)= n(M)$$ means the number of elements of set P = Number of elements of set M.
$$\therefore$$ P and M are equivalent.
Hence, the given statement is true.
M $$\cup$$ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4} then which of the following represent set N?
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0%
{1, 2, 3}
0%
{3, 4, 5, 6}
0%
{2, 5, 6}
0%
{4, 5, 6}
Explanation
We have, M $$\cup$$ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4}
Since, {1, 2, 3} $$\cup$$ {1, 2, 4} = {1, 2, 3, 4} $$\neq$$ M $$\cup$$ N = {1, 2, 3, 4, 5, 6};
{3, 4, 5, 6} $$\cup$$ {1, 2, 4} = {1, 2, 3, 4, 5, 6} = M $$\cup$$ N = {1, 2, 3, 4, 5, 6};
{2, 5, 6} $$\cup$$ {1, 2, 4} = {1, 2, 4, 5, 6} $$\neq$$ M $$\cup$$ N = {1, 2, 3, 4, 5, 6}; and
{4, 5, 6} $$\cup$$ { 1, 2, 4} = {1, 2, 4, 5, 6} $$\neq$$ M $$\cup$$ N = {1, 2, 3, 4, 5, 6}
If $$P \ \subseteq \ M$$, then Which of the following set represent $$P \ \cap \ (P \ \cup \ M)$$ ?
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0%
P
0%
M
0%
P $$\cup$$ M
0%
P' $$\cap$$ M
Find the correct option for the given question.
Which of the following collections is a set?
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0%
Colours of the rainbow
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Tall trees in the school campus
0%
Rich people in the village
0%
Easy examples in the book
Explanation
Since, the colors of the rainbow are well defined such as Violet, Indigo, Blue, Green, Yellow, Orange and Red.
So, the collection of colors of the rainbow is a set.
Examine whether the following statements are true or false:
$$\left\{a,e \right\} \subset \left\{x : x \ is\ a\ vowel\ in\ the\ English\ alphabet \right\}$$
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0%
True
0%
False
Explanation
a
,
e
∈
{
a
,
e
,
i
,
o
,
u
}
In a town of $$10,000$$ families it was found that $$40\%$$ families buy newspaper $$A$$, $$20\%$$ families buy newspaper $$B$$ and $$10\%$$ families buy newspaper $$C$$. $$5\%$$ families buy $$A$$ and $$B$$, $$3\%$$ buy $$B$$ and $$C$$ and $$4\%$$ buy $$A$$ and $$C$$. If $$2\%$$ families buy all the three newspaper, find the number of families which buy (i) $$A$$ only (ii) $$B$$ only (iii) none of $$A, B$$ and $$C$$
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(i) $$3000$$ (ii) $$1800$$ (iii) $$4600$$
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(i) $$3300$$ (ii) $$1400$$ (iii) $$4000$$
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(i) $$3500$$ (ii) $$1600$$ (iii) $$3800$$
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none
Examine whether the following statements are true or false:$$\left\{x : x\ is\ an\ even\ natural\ number\ less\ than\ 6 \right\} \subset \left\{x : x is\ a\ natural\ number\ which\ divides\ 36 \right\}$$
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0%
True
0%
False
Explanation
{
2
,
4
}
∵{2,4}
is a set but no in
{
1
,
2
,
3
,
4
,
6
,
9
,
12
,
18
,
36
}
Examine whether the following statements are true or false:$$\left\{1,2,3 \right\} \subset \left\{1,3,5 \right\}$$
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0%
True
0%
False
Explanation
2
∉
{
1
,
3
,
5
}
Examine whether the following statements are true or false:
$$\left\{a,b \right\} \not \subset \left\{b,c,a \right\}$$
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0%
True
0%
False
Explanation
a
,
e
∈
{
a
,
e
,
i
,
o
,
u
}
In each of the following, determine whether the statements is true or false if it is true prove it if it false given an example.
If $$A \subset B$$ and $$B \subset C$$, then $$A \subset C$$
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0%
True
0%
False
Explanation
Given
A
⊂
B
A⊂B
and
Now we have to prove
A
⊂
C
A⊂C
∴
x
∈
C
∵
B
∈
C
∴x∈C∵B∈C
∀
X
∈
A
⇒
X
∈
C
∴
A
∈
C
Examine the following statements:
{x : x is an even natural number less then 6} $$\subset $$ { x : x is natural number which divide 36 }
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0%
True
0%
False
Explanation
{x : x is an even natural number less then 6} $$\subset $$ { x : x is natural number which divide 36 }
$$ \left \{ 2, 4\right \} \subset \left \{ 1, 2, 3, 4 , 6, 9, 12, 18, 36\right \} $$
Hence, the statement is true
If A = {1, 2, 3}, B = {3, 4, 5}, C = {4, 6}, then
A x $$( B \cup C)$$ =
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0%
{(1, 3) (1, 4) (1, 5) (1, 6) (2, 3) (2, 4) (2, 5) (2, 6) (3,3) (3, 4) (3,5) (3, 6)}
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A x $$(B \cap C)$$
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B x $$(A \cap C)$$
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all the above
Explanation
If A = {1 , 2 , 3} , B = {3 , 4 , 5} , C = {4 , 6}
then, to find A X (B U C)
B U C = combination of unique elements of set B and C.
B U C = {3, 4 , 5, 6}
Using associative property of set:
A X (
B U C) = (A X B ) U ( A X C)
To find cartesian
products:
(i) A X B = {1 , 2 , 3 } X {3 , 4 , 5}
= {(1 , 3), (1 , 4), (1 , 5), (2 , 3), (2 , 4), (2 , 5), (3 , 3), (3 , 4), (3 , 5)}
(i) A X C = {1 , 2 , 3 } X {4 , 6}
= {(1 , 4), (1 , 6), (2 , 4), (2 , 6), (3 , 4), (3 , 6)}
(A X B ) U ( A X C) =
{(1 , 3), (1 , 4), (1 , 5), (2 , 3), (2 , 4), (2 , 5), (3 , 3), (3 , 4), (3 , 5)} U
{(1 , 4), (1 , 6), (2 , 4), (2 , 6), (3 , 4), (3 , 6)}
Ans =
{(1 , 3), (1 , 4), (1 , 5),
(1 , 6),
(2 , 3), (2 , 4), (2 , 5)
, (2 , 6)
, (3 , 3), (3 , 4), (3 , 5),
(3 , 6)
}
Which of the following statements is true (if N, W and I are sets of Natural, Whole and
Integer numbers respectively ?
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$$N\, \subset \, W\, \subset \, I$$
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$$I\, \subset \, N\, \subset \, W$$
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$$W\, \subset \, N\, \subset \, I$$
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$$I\, \subset \, W\, \subset \, N$$
S = {1, 2, 3, 5, 8, 13, 21, 34 }. Find $$\displaystyle \sum $$ max (A), where the sum is taken over all 28 elements subsets A to S.
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0%
844
0%
480
0%
484
0%
488
There are 6 boxes numbered 1, 2, ...Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:
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0%
5
0%
21
0%
33
0%
60
The set $$\{x/| x L|< K\}$$ is the same for all $$K > 0$$ and for all L, as
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0%
$$\{x/0 < x < L + K\}$$
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$$\{x/L K < x < L + K\}$$
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$$\{x/|L K| < x <|L + K|\}$$
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$$\{x/|L x| > K\}$$
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$$\{x/ K < x < L\}$$
If $$S$$ represents the set of all real numbers $$x$$ such that $$1\le x \le 3$$ and $$T$$ represents the set of all real numbers $$x$$ such that $$2 \le x \le 5$$, the set represented by $$S \cap T$$ is
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0%
$$2 \le x \le 3$$
0%
$$1 \le x \le 5$$
0%
$$x \le 5 $$
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$$x\ge 5$$
0%
none of these
Explanation
$$S\cap T$$(common part of S and T) = $$2\leq x\leq 3$$
Suman is given an aptitude test containing 80 problems, each carrying I mark to be tackled in 60 minutes. The problems are of 2 types; the easy ones and the difficult ones. Suman can solve the easy problems in half a minute each and the difficult ones in 2 minutes each. (The two type of problems alternate in the test). Before solving a problem, Suman must spend one-fourth of a minute for reading it. What is the maximum score that Suman can get if he solves all the problems that he attempts?
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0%
30
0%
40
0%
53
0%
60
Explanation
$$Assuming\quad that\quad all\quad the\quad problems\quad suman\quad solves\quad are\quad right,\\ marks\quad earned\quad per\quad minute\quad solving\quad easy\quad problem=\dfrac { 1 }{ 3/4 } =\dfrac { 4 }{ 3 } \\ marks\quad earned\quad per\quad minute\quad solving\quad difficult\quad problem=\frac { 1 }{ 9/4 } =\frac { 4 }{ 9 } \\ since\quad we\quad have\quad to\quad maximize\quad the\quad marks,\quad we\quad will\quad solve\quad all\quad the\quad easy\quad problems\quad first.\\ marks\quad earned\quad by\quad solving\quad easy\quad problems=40,\\ time\quad spent=\frac { 3 }{ 4 } *40=30\quad minutes\\ time\quad left\quad for\quad difficult\quad problems=30\quad minutes,\\ problems\quad solved=\frac { 30 }{ 9/4 } =13,\\ marks\quad earned=13,\\ total\quad marks=40+13=53$$
Union set is defined as
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a collection of sets is the set of all elements in the collection
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It is one of the fundamental operations through which sets can be combined and related to each other.
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Set whole each element is an element of all the present set
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None of these
The dual of $$-p\wedge (q\vee \sim r)$$ is
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$$p\vee (\sim q \wedge r)$$
0%
$$\sim p\vee (q\wedge r)$$
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$$p\wedge (q\wedge r)$$
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$$\sim q\vee(q\wedge \sim r)$$
If $$20$$% of three subsets (i.e., subsets containing exactly three elements) of the set $$A = \left \{a_{1}, a_{2}, ...., a_{n}\right \}$$ contain $$a_{2}$$, then the value of $$n$$ is
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0%
$$15$$
0%
$$16$$
0%
$$17$$
0%
$$18$$
Suppose $${ A }_{ 1 },{ A }_{ 2 },,{A }_{ 30 }$$ are thirty sets each having $$5$$ elements and $${ B }_{ 1 },{ B }_{ 2 },..,{B}_{ n }$$ are $$n$$ sets each with $$3$$ elements, let $$\displaystyle \bigcup _{ i=1 }^{ 30 }{ { A }_{ i } } =\bigcup _{ j=1 }^{ n }{ { B }_{ j } =S}$$ and each element of $$S$$ belongs to exactly $$10$$ of the $${A}_{i}s$$ and exactly $$9$$ of the $${B}_{j}s.$$ Then $$n$$ is equal to
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0%
$$15$$
0%
$$3$$
0%
$$45$$
0%
$$None\ of\ these$$
Explanation
$$(c)$$.
Since each $$A_{ i }$$ has $$5$$ elements, we have
$$\overset { 30 }{ \underset { i=1 }{ \Sigma } } \ n\left( { A }_{ i } \right) =5\times 30=150$$. . . $$(1)$$
Let $$S$$ consist of $$m$$ distinct elements. Since each elements of $$S$$ belongs to exactly $$12$$ of the $$A_{ i }$$ $$s$$ we also have
$$\overset { 30 }{ \underset { i=1 }{ \Sigma } } \quad n\left( { A }_{ i } \right) =10m$$. . . $$(2)$$
Hence from $$(1)$$ and $$(2)$$, $$10m=150$$ or $$m=15$$. Again since each $$B_{ i }$$ has $$3$$ elements and each element of $$S$$ belongs to exactly $$9$$ of the $$B_{ j }$$ $$s$$ we have
$$\overset { 30 }{ \underset { j=1 }{ \Sigma } } \quad n\left( { B }_{ j } \right) =3n$$ and $$\overset { 30 }{ \underset { j=1 }{ \Sigma } } \quad n\left( { B }_{ j } \right) =9m$$
It follows that $$3n=9m=9\times 15$$
. . . $$[\therefore m=15]$$
This gives $$n=45$$.
All the permissible values of $$b$$, if $$a=0$$ and $${S}_{2}$$ is a subset of $$\left( 0,\pi \right) $$
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0%
$$b\in \left( -n\pi ,2n\pi \right) ;\in Z$$
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$$b\in \left( -n\pi ,2\pi -n\pi \right) ;\in Z$$
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$$b\in \left( -n\pi ,n\pi \right) ;\in Z$$
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none of these
If $$S_k=\begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix}$$, k$$\in N^+$$, where N is the set of all natural numbers, then $$(S_2)^n(S_k)^{-1}$$ for n$$\in$$N is?
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0%
$$S_{2n-k}$$
0%
$$S_{2n+k-1}$$
0%
$$S_{2^n+k-1}$$
0%
$$S_{2^n+k+1}$$
$$13$$ Planar Surfaces, each of area $$1\ m^{2}$$ are such that their total area is $$7\ m^{2}$$. If no three of more of these surfaces have any region in common; Then, the overlap of some Two of These surfaces has an Area.
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Not less than $$\dfrac{7}{13} m^{2}$$
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Not less than $$\dfrac{2}{13} m^{2}$$
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Not less than $$\dfrac{1}{13} m^{2}$$
0%
Not Less than $$\dfrac{1}{17} m^{2}$$
Explanation
Assume are of overlap of any $$2$$ surfaces is less than $$\dfrac{1}{13}m^2$$
Let us ignore the unit of area in our notation.
We know that, $$Area(A_{1}\cup A_{2}) = Area(A_{1}) + Area(A_{2}) - Area(A_{1}\cap A_{2})$$
But, $$Area (A_{1}) = Area (A_{2}) = 1.$$ (as given in question)
Also, $$Area(A_{1}\cap A_{2}) < \dfrac {1}{13}$$ (as per our assumption)
Hence, $$Area(A_{1}\cup A_{2}) > 1 + 1 - \dfrac {1}{13}$$
$$Area(A_{1}\cup A_{2}) > 2 - \dfrac {1}{13}$$
Similarly,
$$Area (A_{1}\cup A_{2}\cup A_{3}) = Area (A_{1}) + Area(A_{2}) + Area(A_{3})$$
$$ - Area(A_{1}\cap A_{2})- Area (A_{1}\cap A_{3}) - Area(A_{2} \cap A_{3})$$
$$ + Area (A_{1}\cap A_{2}\cap A_{3})$$
But $$Area(A_{1}\cap A_{2}\cap A_{3}) = 0$$ (As per the condition in Question)
Thus, $$Area(A_{1}\cup A_{2}\cup A_{3}) > 1 + 1 + 1 - \dfrac {1}{13} - \dfrac {1}{13} - \dfrac {1}{13} + 0$$.
i.e. $$Area(A_{1} \cup A_{2} \cup A_{3}) > 3 - \dfrac {3}{13}$$.
Similarly,
$$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) = Area(A_{1}) + Area (A_{2}) + Area(A_{3}) + Area(A_{4})$$
$$- Area(A_{1}\cap A_{2}) - Area(A_{1}\cap A_{3}) - Area (A_{1} \cap A_{4})$$
$$- Area(A_{2} \cap A_{3}) - Area(A_{2} \cap A_{4}) - Area (A_{3} \cap A_{4})$$
$$ + Area(A_{1}\cap A_{2}\cap A_{3})+ Area(A_{1}\cap A_{2}\cap A_{4})$$
$$ + Area (A_{2}\cap A_{3}\cap A_{4}) - Area (A_{1}\cap A_{2}\cap A_{3}\cap A_{4})$$
But $$Area(A_{1}\cap A_{2}\cap A_{3}\cap A_{4}) = 0$$
$$Area (A_{1} \cap A_{2}\cap A_{3}) = 0$$, etc. (as per the condition in Question)
So, $$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) > 1 + 1 + 1 + 1 - 6\left (\dfrac {1}{13}\right ) + 3(0) - 0$$.
i.e. $$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) > 4 - \dfrac {6}{13}$$
In general,
$$Area(A_{1}\cup A_{2}\cup A_{3}\cup ...\cup A_{n}) > n - \dfrac {^{n}C_{2}}{13}$$
Put $$n = 13$$;
$$Area(A_{1}\cup A_{2}\cup A_{3}\cup ....\cup A_{13}) > 13 - \dfrac {^{13}C_{2}}{13}$$
That is $$Area(A_{1}\cup A_{2}\cup A_{3}\cup ....\cup A_{13}) > 7$$
Hence, our assumption was wrong.
So, Area of overlap of any $$2$$ surfaces is not less than $$\dfrac {1}{13}$$.
State which of the following is total number of reflexive relations form set $$A = \left \{a, b, c\right \}$$ to set $$B = \left \{d, e\right \}$$ is
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0%
$$2^{6}$$
0%
$$2^{8}$$
0%
$$2^{4}$$
0%
$$2^{15}$$
Explanation
Number of reflexive relation from$$A=[a,b,c]$$ and $$B=[d,e]$$is
$$=2^{n^2-n}=2^{3^2-3}=2^{9-3}=2^6$$
Sets $$A$$ and $$B$$ have $$5$$ and $$6$$ elements respectively and $$\left( A\triangle B \right) =C$$ then the number of elements in set $$\left( A-\left( B\triangle C \right) \right)$$ is
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0%
$$5$$
0%
$$6$$$
0%
$$0$$
0%
$$4$$
Suppose $$A_1 , A_2,... A_{30}$$ are thirty sets each having 5 elements and $$B_1, B_2,..., B_n$$ are n sets each with 3 elements , let $$\underset{i = 1}{\overset{30}{\cup}} A_i = \underset{j = 1}{\overset{n}{\cup}} B_j = S$$ and each element of S belongs to exactly 10 of the $$A_i's$$ and exactly 9 of the $$B_j'S$$. then n is equal to
Report Question
0%
15
0%
3
0%
45
0%
35
Explanation
If elements are not repeated then the number of elements in $$A_1\cup A_2\cup A_3\cup A_{30}$$ is $$30\times 5$$
But each elements is used 10 times
so, $$S=\cfrac{30\times 5}{10}=15$$
If the elements in $$B_1,B_2,.....,B_n$$ are not repeated.
then total number of element in
$$B_1\cup B_2\cup B_3.....\cup B_n$$ is 3n
but each elements is repeated 9 times
so,
$$S=\cfrac{3n}{9}=15=\cfrac{3n}{9}$$
$$n=45$$
Let A and B be 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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0%
$$275$$
0%
$$510$$
0%
$$219$$
0%
$$256$$
Explanation
Set A has $$4$$ elements
Set B has $$2$$ elements
Number of elements in set $$\left( A\times B\right)=4\times 2=8$$
Total number of subsets of $$\left( A\times B\right) =2^8=256$$
Number of subsets having $$0$$ elements $$=^{ 8 }{ { C }_{ 0 } }=1$$
Number of subsets having $$1$$ elements each $$=^{ 8 }{ { C }_{ 1 } }=8$$
Number of subsets having $$2$$ elements each $$=^{ 8 }{ { C }_{ 2 } }=\dfrac{8!}{2!6!}=28$$
Number of subsets having atleast $$3$$ elements $$=256-1-8-28$$
$$=256-37$$
$$=219$$
Hence, the answer is $$219.$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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