We use partitions to enumerate sets.
When counting the number of elements in a set, it is sometimes helpful to break the set into distinct pieces and count the number of elements in each piece.
We will denote the number of elements in a set by . The main result is that the number of elements in a set is the sum of the elements in each subset of the partition: where is a partition of .
We have seen that the total number of outcomes of the two dice when recorded as (red, green) is 36. How many of these 36 possible outcomes produce a sum of 10 or higher (10, 11 or 12)? If we let be the set of all outcomes whose sum is greater than or equal to 10, then we can partition into three subsets: and where the subscript identifies the sum. The set contains the outcomes and , so . Verify that and .
Finally, since is a partition of , we have Thus there are 6 ways to get a sum of at least 10 when rolling a pair of 6-sided dice.
The number of outcomes whose sum is at most 5 is .
The number of outcomes whose sum is prime is .
The number of outcomes whose sum is odd is .
If we denote by the set of all possible outfits Steph can wear, then we can partition into those outfits with a hat, , and those without a hat, . The number of outfits with a hat is and the number of outfits with no hat is . Thus, the total number of ways for Steph to dress is Note that we could have counted this slightly differently, by including “no hat” as a option for the hat: .
If we denote by the set of all possible acceptable slices of pizza, then we can partition into the slices that have pepperoni, and those that do not, . The number of slices of each is 3 and hence, the total number of possible slices Sam can order is
If is the set of all possible genders of their children, then we can partition according to the number of children. Let be the number of ways for them to have 0 children. Then because the only way to have 0 children is to have 0 boys and 0 girls. Let be the number of ways for them to have 1 child. Then because they can have 1 boy and 0 girls or 1 girl and 0 boys. Let be the number of ways for them to have 2 children. Then because they can have 2 boys and 0 girls, 2 girls and 0 boys or 1 of each. Thus the total number of combinations of boys and girls that they can have is