recategorized by
4,525 views
4 votes
4 votes

The number of elements in the power set of $\{\{1,2\},\{2,1,1\},\{2,1,1,2\}\}$ is

  1. $3$
     
  2. $8$
     
  3. $4$
     
  4. $2$
recategorized by

3 Answers

Best answer
25 votes
25 votes

Consider this: {1} & {1,1} are the same set. {Since, order & repetition of elements in the set won't make it a new set}

Here, {1,2}, {2,1,1} & {2,1,1,2} are same & so the number of elements in A will be 1

Number of elements in Power-set(A) = 2= 2

selected by
5 votes
5 votes
no. of elements in power set of A = $2^{|A|}$

|A| = 3, therefore $|power-set(A)| = 2^{3} =8$
5 votes
5 votes
If the given set is a multiset. Answer is 8.

If it is a simple set. Answer is 2.
Answer:

Related questions

6 votes
6 votes
2 answers
1
gatecse asked Dec 17, 2017
2,643 views
The function $f:[0,3]\rightarrow [1,29]$ defined by $f(x)=2x^{3}-15x^{2}+36x+1$ isinjective and surjectivesurjective but not injectiveinjective but not surjectiveneither ...
4 votes
4 votes
2 answers
2
gatecse asked Dec 17, 2017
3,135 views
Let $f(x)=\log|x|$ and $g(x) =\sin x$. If $A$ is the range of $f(g(x))$ and $B$ is the range of $g(f(x))$ then $A\cap B$ is$[-1,0]$$[-1,0)$$[-\infty ,0]$$[-\infty ,1]$
3 votes
3 votes
1 answer
3
gatecse asked Dec 17, 2017
1,770 views
If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log \mid x \mid+\beta x^2+x$ then$\alpha=-6,\beta=\dfrac{-1}{2}$ $\alpha=2,\beta=\dfrac{-1}{2}$ $\alpha=2,\beta=\d...
12 votes
12 votes
2 answers
4