Question

Ok ! Let k = {1,2,..,n} . How many subsets are there for k

Answer 1

\[\large k(n)\]\[ k(1)=\{\},\{1\}\]\[k(2)=\{\},\{1\},\{2\},\{1,2\}\]\[\small k(3)=\{\},\{1\},\{2\}, \{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\]\[\tiny k(4)=\{\},\{1\},\{2\}, \{3\}, \{4\}, \{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\]

Answer 2

\[ 2^n=(1+1)^n=\sum_{j=0}^n\binom{n}{j} \]

Answer 3

n^2

Answer 4

the correct answer was 2^n

Answer 5

Can any1 help me ?

Answer 6

This is referred to as the power set of k. the notation for it is \(2^k.\) The order of \(2^k\) is \(2^{|k|}\), which is \(2^n\).

Answer 7

@nbouscal am I right that the 2^n part comes from a binary observation of selecting or not selecting a particular number to be a member of the subset? or am I just talking gibberish?

Answer 8

Yes, that is definitely a way of viewing the notion.

Answer 9

thanks!

Answer 10

each possible element ( and the possibility for zero elements) is a degree of freedom for the system , each is either present or not present in each set

Answer 11

that's what I meant by "binary"
either each element is there or not; 2 choices

Answer 12

binary is a good word to use here , each number is like a switch

Answer 13

@mathslover did we help you understand, or just talk among ourselves ?

Answer 14

Hmn @TuringTest and @UnkleRhaukus and @nbouscal I am going to post what i did ! can u check that out ?

Answer 15

is this right ? @TuringTest @nbouscal @UnkleRhaukus

Answer 16

any comments please ? m i right @TuringTest