Ordered pair of subsets

it is 2^5.because each element has only two choices. either it can go to a or to b .it cant go to both becoz a intersec b is phi it cant go to none becoz a u b is A

It is given that $A \cap B = \phi$ . This means that $A$ and $B$ have no elements in common. Also, $A \cup B = U$ . This means that elements of $B$ are those which are not in $A$ . So, for each set $A$ , there is only one set $B$ .

When there are no elements in $A$ and 5 elements in $B$ , only one ordered pair is possible, which is $A = \phi$ and $B = {{1,2,3,4,5}}$ .

Similarly, when there is 1 element in $A$ and 4 elements in $B$ , 5 ordered pairs are possible. When there are 2 elements in $A$ and 3 elements in $B$ , 10 ordered pairs are possible. When there are 3 elements in $A$ and 2 elements in $B$ , 10 ordered pairs are possible. When there are 4 elements in $A$ and 1 elements in $B$ , 5 ordered pairs are possible.

Lastly, the reverse of the first case is also possible, where $B = \phi$ and $A = {1,2,3,4,5}$ .

So, in all there are $1 + 5 + 10 + 10 + 5 + 1 = \boxed{32}$ cases.

Such sets are called 'Partition Sets', mutually exclusive and exhaustive. To calculate total no. of partition sets, ${ 2 }^{ n }$ .

A simple way to understand using permutations.

Let the two sets be Set 0 & Set 1 $\underline { } \underline { } \underline { } \underline { } \underline { }$

The spaces denoting $1,2,3,4,5.$ $$

where $\underline { 0 } \underline { 0 } \underline { 0 } \underline { 1 } \underline { 1 }$ would mean

$Set 0 = 1, 2, 3$

$Set 1 = 4, 5$

Then taking 1,1,1,1,1 , 1,1,1,1,0 , 1,1,1,0,0 & so on till 0,0,0,0,0

Permutations would be

$\frac { 5! }{ 0!.5! } + \frac { 5! }{ 1!4! } + \frac { 5! }{ 2!3! } + \frac { 5! }{ 3!2! } + \frac { 5! }{ 4!1! } + \frac { 5! }{ 5!0! }$

which would, according to Binomial Theorem be ${ (1+1) }^{ 5 }$

First we realize it is a "n choose r" idea. It is a summation of 5 choose r with r from 0 to 5. Using the binomial theorem we know sum of n choose r = 2^n. Therefore the answer is 2^5 = 32

There are 32 subsets, so if A is any one of them, let B be the complement of A. This exhausts all possibilities. Ed Gray

total no of ways is 2^5=32 as it is no of ways each number can be distribute among two set

The answer is 2^n, where n is the number of numbers in U.