Problem Solving With Combinations

In this section, we will look at scenarios in which you wish to determine how many alternative combinations of any size you may select from a specified number of elements, some of which might be identical. Any collection of unique objects can be subjected to the same reasoning.

The total number of combinations containing at least one item chosen from a group of \(n\) distinct items can be represented algebraically as:

\(2^n-1\)

Remember that combinations are subsets of the group of n objects. A null set is a set that has no elements. Therefore, the null set is one of \(2^n\) subsets of a set with \(n\) unique members.


Example

In how many ways can a team with at least one player be selected from a squad of eight players?

The squad can select anywhere from 1 to 8 players, so \( n = 8 \). Since the team must have at least one player, we use the formula that excludes the null set:

\(\text{Combinations} = 2^n - 1\)

\(= 2^8 - 1\)

\(= 256 - 1\)

\(= 255\)

Therefore, we can determine there are 255 ways to form a team of at least one player from an eight-member squad.


Example

A picnic planner can select three types of sandwiches, two types of drinks, and two types of desserts to include in a picnic basket. If there has to be at least one selection, how many distinct picnic baskets can be made?

In order to calculate the total number of combinations, we can use the formula for independent choices where each type of item (sandwiches, drinks, desserts) can either be included or not:

\(2^3\) ways for sandwiches, \(2^2\) for drinks, \(2^2\) for desserts.

Next, we can multiply these possibilities to get the total combinations when choosing freely from any food item:

\(\text{Combinations} = (2^3 \times 2^2 \times 2^2)\)

\(= 8 \times 4 \times 4\)

\(= 128\)

Since the basket must contain at least one food item, we can subtract the one scenario where no item is chosen:

\(\text{Combinations} = 128-1\)

\(= 127\)

Therefore, we can detemrmine there are 127 different ways to assemble a picnic basket with at least one item. This confirms the total number of unique picnic setups excluding the null set.


When preparing a bouquet, a florist can choose from seven roses, six carnations, and four chrysanthemums. How many bouquets can the florist make if the bouquet must have at least one flower?

To determine the number of possible bouquets from seven roses, six carnations, and four chrysanthemums, ensuring at least one flower is included, follow this calculation:

First, we can calculate the total number of combinations for each flower type independently, considering each can either be included or not:

\(2^7\) ways for roses, \(2^6\) for carnations, and \(2^4\) for chrysanthemums.

Next, we can multiply these possibilities to get the total combinations when choosing freely from any flowers:

\(\text{Combinations} = 2^7 \times 2^6 \times 2^4\)

\(= 128 \times 64 \times 16\)

\(= 131,072\)

Since the bouquet must contain at least one flower, we can subtract the one scenario where no flower is chosen:

\(\text{Combinations} = 131,072 - 1\)

\(= 131,071\)

Therefore, the florist can create 131,071 different bouquets when at least one flower is used.





Understanding Combinations with Similar Items

A specific method is needed for computing combinations, especially when some of the pieces are similar. Since selecting the same item in various sequences shouldn't be considered separate selections, identical items within a collection make the simple application of combinations more difficult.

The formula \((p + 1)(q + 1)(r + 1) \ldots - 1\) is used in scenarios where combinations involve categories of identical items, such as multiple types of snacks or books. Each term like \(p+1\) or \(q+1\) includes the option to select zero items from that category, covering all potential choices from selecting none to all items available.

This formula determines the overall number of ways to combine elements, even if we do not choose any items from specific categories. We subtract one from the total to ensure that we do not include the option where no items are selected at all. This ensures that each option we count contains at least one item. The formula may be adjusted dependent on how many sorts of objects there are, making it highly versatile for a number of scenarios where you're grouping similar items together.


Example

A teacher is preparing snack packs for a school party and has 3 varieties of snacks: 3 packets of cookies, 4 packets of gummy bears, and 2 packets of pretzels. How many different types of snack packs can be created, assuming that each pack contains at least one snack?

The teacher has 4 choices for each type of cookie snack (\(0\) to \(3\) packets), \(5\) choices for each type of gummy bear snack (\(0\) to \(4\) packets), and 3 choices for each type of pretzel snack (\(0\) to \(2\) packets).

Using the formula \((p + 1)(q + 1)(r + 1) - 1\), we can calculate the number of different snack pack combinations:

\(\text{Combinations} = (3+1)(4+1)(2+1) - 1\)

\(= 4 \times 5 \times 3 - 1\)

\(= 60 - 1\)

\(= 59\)

Therefore, we can determine there are 59 different ways to make the snack packs.


A librarian is setting up a display table at the library with a selection of 5 historical novels, 3 science fiction novels, and 4 mystery novels. How many different arrangements can the librarian make for the display, featuring at least one book but no more than one from each genre?

The librarian has 2 choices for showing historical novels (\(0\) or \(1\) book), 2 choices for showing science fiction novels (\(0\) or \(1\) book), and 2 choices for showing mystery novels (\(0\) or \(1\) book).

Using the formula \((p + 1)(q + 1)(r + 1) - 1\), we can calculate the total number of unique book display setups:

\(\text{Combinations} = (1+1)(1+1)(1+1) - 1\)

\(= 2 \times 2 \times 2 - 1\)

\(= 8 - 1\)

\(= 7\)

Therefore, we can determine there are 7 different ways to set up the book display.