So What's the Deal With Power Sets?

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A Power Set...

  • Is the set of all possible subsets
  • Includes the empty set (X = 0 members in the subset)
  • Doesn't care about order
    • (i.e., order doesn't matter)
  • Is the sum of the combinations of all subsets

A Power Set...

  • Is not a permutation. Order does not matter in a power set.
    • (In fact, for mathematicians working in this area, calling something a set specifically means the collection is unordered. Ordered collections are called lists or ​strings.)
  • Is not a combination. Combinations concern themselves with one specific value of X. Power sets consider all values of X. So the number of items in the power set is the sum of all individual combinations.

An Example

My mother bought me a scarf, a pair of mittens, and a hat to wear this winter.

 

I can wear

  • none of it (0 items),
  • 1 item,
  • 2 items,
  • or all 3 items.

 

How many different ways can I wear or not wear these 3 items? That is, how many different subsets are there?

Let's Count the Ways

  • Subsets of size = 0 (i.e., I wear nothing):
    • [  ] = 1 subset
  • Subsets of size = 1:
    • [hat]  [mittens]  [scarf] = 3 subsets
  • Subsets of size = 2:
    • [hat mittens]  [hat scarf]  [mittens scarf] = 3 subsets
  • Subsets of size = 3:
    • [hat mittens scarf] = 1 subset

So there are 1 + 3 + 3 + 1 = 8 possible subsets

Let's Count the Ways Using      

 

Subsets of 0:

 

Subsets of 1:

 

Subsets of 2:

 

Subsets of 3:

 

So there are 1 + 3 + 3 + 1 = 8 possible subsets

_nC_X
_3C_0 = \dfrac{3!}{0!3!} = 1
_3C_1 = \dfrac{3!}{1!2!} = 3
_3C_2 = \dfrac{3!}{2!1!} = 3
_3C_3 = \dfrac{3!}{3!0!} = 1

Like I said, the number of items in the power set is the sum of all combinations, X = 0 to n.

Let's Count the Ways

Using Binary Coding

Let's code an item 1 if I wear it and 0 if I don't wear it (i.e., 1 if the item is in the subset, 0 if it's not).

The pattern is: [hat mittens scarf].

  • [  ] = [0 0 0]
  • [hat] = [1 0 0]
  • [mittens] = [0 1 0]
  • [scarf] = [0 0 1]
  • [hat mittens] = [1 1 0]
  • [hat scarf] = [1 0 1]
  • [mittens scarf] = [0 1 1]
  • [hat mittens scarf] = [1 1 1]

So you can see, to find the number of possible subsets, you have to read the largest subset (the subset with all 1s in it) as a binary number. In this case, that's 111, which is 7 in decimal form. Adding the empty set, [  ], makes 8.

Let's Count the Ways Using the Quickest Method

The simplest way to count subsets in a power set is to form the largest binary number (111), plus 1 for the empty set, as an exponent with base 2:

2^n = 2^3 = 8,

where n is the number of items in the mother set. 

 

That's 2 possible values (0 or 1) for hat, 2 for mittens, and 2 for scarf. So there are 2 possibilities of hat AND 2 more possibilities  of mittens AND 2 more possibilities of scarf.

Combined, that's 2 x 2 x 2 = 8 total possibilites.

OK, One More Example

I want to take my Sunday school class on a trip to hear AC/DC in concert. There are 6 students in the class. I may have 0 come to the concert, 1, 2, 3, 4, 5, or all 6.

How many different subsets are possible?

2^n = 2^6 = 64

So there are 64 possible combinations of students that could go to the concert.

Fin

1. Probability - Counting Power Sets

By smilinjoe

1. Probability - Counting Power Sets

The set of all possible subsets one could form from the mother set.

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