Question

How many subsets are there of a set consisting of n elements?

Answer 1

\(2^n \)
are interesting in proving it ?

Answer 2

The answer is right but can you please show the work?

Answer 3

Consider a subset of the given set.

Each element in the given set can either exist in the subset or not : 2 choices
Since there are \(n\) elements, the total possible combinations of elements would be
\[2\times 2\times 2\times \cdots \text{(n times)} = 2^n\]

Answer 4

Another way to think of it:
\[\begin{array}{c|c|c|c|c}
\text{Number of elements}&\text{Ways to have them}\\
\hline
0&1=\binom n0\\
1&n=\binom n1\\
2&\frac{n(n-1)}{2}=\binom n2\\
\vdots&\vdots\\
n-1&\binom n{n-1}\\
n&\binom nn
\end{array}\]
So the total number of ways is
\[\sum_{k=0}^n \binom nk\]
then refer to the binomial theorem.

Answer 5

yeah and there is a cool thing about this formula , which is if u have a set
A={ }
which mean 0 elements in A , the number of subsets is 2^0=1 :)
so the only subset of A is \(\varnothing \) and power set of A would be denoted by
\(P(A)=\{\varnothing \}\)

Answer 6

use `\mathcal{P}` for powerset ikram

Answer 7

\( \mathcal{P} ardon \Huge ت\)

Answer 8

\[\mathcal{P}(A) = \sum\limits_{k=0}^{n}\binom{n}{k} = \sum\limits_{k=0}^{n}\binom{n}{k}1^{n-k}1^k = (1+1)^n \]

Answer 9

well hmm if u mean by that formula number of sets in the power set then ok
but = thingy xD

Answer 10

ok i got what ur saying

Answer 11

Thank you so much guys! I really appreciated it!

Answer 12

\[ |\mathcal{P}(A) |= \sum\limits_{k=0}^{n}\binom{n}{k} = \sum\limits_{k=0}^{n}\binom{n}{k}1^{n-k}1^k = (1+1)^n \]

Answer 13

haha roger , also u can call it order of \( \mathcal{P} \) set