Combinations

Fundamental Counting Principle: the principle for determining the number of ways two or more operations can be performed together
permutation: an arrangement of all or part of a set of objects, where the order of the arrangement is important
If you have gone shopping for clothes with friends or family, you have probably had to wait patiently until they made a decision as to what to purchase. To help customers make up their minds, salesclerks lay out a couple of pairs of pants and several shirts. This way, customers can quickly make comparisons and select the preferred combination.
In mathematics, a selection from a set of objects is also called a combination. When a selection is made, the order in which the elements are chosen is not important. The same would be true, for example, when choosing a shirt and pants combination. It doesn’t matter which item you buy first; what is important is the final look of the outfit.
permutation: an arrangement of all or part of a set of objects, where the order of the arrangement is important. Form is _{n}P_{r}.
Note: _{n}P_{r} = _{n}C_{r} × r!.
There are r! times as many permutations as combinations because there are r! ways of arranging the the r objects once they are chosen.
The permutation _{n}P_{r} represents two tasks: selecting r objects from a set of n objects first, then arranging the r objects selected. Using the Fundamental Counting Principle,
Equivalent Combinations: If two combinations have the same n value and the sum of their rvalues is equal to n, then the two combinations are equal. Examples:
When you choose 2 objects from 10, there are are 8 left. This is the same as selecting 8 objects from 10 and having 2 left.
Use factorial notation to show that _{10}C_{2} = _{10}C_{8}
A customer is trying to decide between 2 different pairs of pants and 6 different shirts. How many combinations of 1 pair of pants and 1 shirt are possible?

SolutionUse a tree diagram to illustrate the possible selections. In the diagram, use P_{1} and P_{2} to represent the pants and S_{1}, S_{2}, S_{3}, S_{4}, S_{5}, and S_{6} to represent the shirts. Using the Fundamental Counting Principle, 2 × 6 = 12 outfits are possible. 
Determining the number of ways a selection from a set can be made when the order does matter.
A tray contains 4 different cookies.
a. How many ways can you select and arrange 3 cookies from the tray? 
b. How many ways can you select 3 cookies from the tray if order doesn’t matter?  c. How are the answers to questions a. and b. related? 
SolutionOrder is important. Therefore, this is a permutation of 4 objects taken 3 at a time. There are 24 ways of selecting and arranging three cookies from the tray. 
SolutionLet the set {A, B, C, D} represent the cookies on the tray. The threecookie combinations are {A, B, C}, {A, B, D}, {A, C, D}, and {B, C, D}. There are 4 ways of selecting three cookies from the tray if order does not matter. 
SolutionThe symbol used to represent the number of combinations is _{4}C_{3}, which is read “4 choose 3.” This implies, “From 4, choose 3.” Here, _{4}C_{3} = 4.
The permutation _{n}P_{r} represents two tasks: selecting r objects from a set of n objects first, then arranging the r objects selected. Using the Fundamental Counting Principle, 
How many different 3member committees can be selected from a class of 20 students?
Order is not important; so, determine the number of combinations.
Method 1: Using Pencil and Paper

Method 2: Using a Graphing Calculator

There are 1140 different committees that can be selected. 
There are 1140 different committees that can be selected 
A cookie jar contains 5 cookies. How many ways can a child choose all 5 cookies if order doesn’t matter?

A cookie jar contains 5 cookies. A child is offered a cookie, but refuses. How many ways can this be done?

SolutionBecause the order doesn’t matter, determine the number of combinations. If order is not important, the child can choose all five cookies only 1 way! 
SolutionThere is only 1 way the child can reject the cookie. This is done simply by saying, “No.” 
A Pure Mathematics 30 class has 10 boys and 12 girls. The teacher wants to form a committee of 3 students to plan the yearend picnic. Determine the number of committees possible if
There are no restrictions

There are no boys on the committee  There must be at least one boy on the committee 
Solution
Without restrictions, there are 1540 committees possible. 
SolutionBecause there are no boys on the committee, you must select committee members from the 12 girls. There are 220 possible committees of only girls. 
SolutionMethod 1: Considering All Possible Cases
Method 2: Using an Indirect Approach

How many diagonals are there in a regular octagon?

How many diagonals are in a regular polygon with 20 sides?  How many triangles can be formed by joining the vertices of a regular polygon with 20 sides? 
SolutionA regular octagon has 8 sides and 8 vertices. Choose any two vertices and join them. The line segment formed is either a side or a diagonal. The total number of ways of joining the 8 vertices is _{8}C_{2}.

Solution 
Solution 
If 3(_{n}C_{2}) = 30 solve for n. Remember that for _{n}C_{r}, n r.

If _{n}C_{3} = 3(_{n}P_{2}) solve for n. Remember that for _{n}C_{r}, n r.

SolutionBecause _{n}C_{2} is only defined when n 2, where n is a whole number, n = 5. Remember that for _{n}C_{r}, n r. 
Solution 
All the possible twonumber permutations and combinations from {1, 2, 3, 4, 5}.

All the possible threenumber permutations and combinations from {1, 2, 3, 4, 5}.

SolutionThe number of permutations of two numbers from {1, 2, 3, 4, 5} with no repetitions is Each of the combinations in this question can be arranged in two ways. The second list has double the number of pairs compared to the first list. Notice that 2 = 2! 
SolutionThe number of permutations of two numbers from {1, 2, 3, 4, 5} with no repetitions is _{5}P_{3} = 5 × 4 × 3 = 60 There are 3! or 6 times as many permutations as combinations. Order is important with permutations, but not with combinations. Notice that 6 = 3! _{5}C_{3} = _{5}P_{3} / 3! = 60 / 6 = 10 
There are six colours: red, white, blue, green, yellow, and black. Find the number of twocolour combinations.

There are six colours: red, white, blue, green, yellow, and black. Find the number of threecolour combinations.

SolutionThere are six colours: red, white, blue, green, yellow, and black.

SolutionThere are six colours: red, white, blue, green, yellow, and black.
There are 20 threecolour combinations. 
The history of combinatorics and probability is imbedded in games of chance; therefore, many examples and questions on the diploma examination involve a standard deck of cards. If you are not familiar with what a standard deck of cards is made up of, read on.
A standard deck consists of 52 cards divided into 4 suits: clubs, spades, hearts, and diamonds.
Each suit has 13 cards: 1 ace, 9 numbered cards (from 2 to 10), and 3 face cards (jack, queen, and king).
How many 13 card hands can be made from a deck of 52 cards?

Bridge hands have 13 card hands. How many bridge hands can be made with 4 aces?  How many bridge hands can be made with no aces? 
Solution 
Solution

Solution
There are approximately 1.93 × 10^{11} different hands with no aces. 
How many 5 card hands can be made from a deck of 52 cards?

How many 5 card hands with only red cards can be made from a deck of 52 cards?  How many 5 card hands can be made of the same suit? 
Solution
There are 2 598 960 different fivecard hands.
There are 52 ÷ 2 = 26 red cards in a deck of cards. 
SolutionThere are 65 780 different fivecard hands containing only red cards

SolutionThere are 4 suits in a standard deck of cards; therefore, there are 52 ÷ 4 = 13 cards in each suit.
There are 5148 different fivecard hands containing all cards of the same suit. 
How many distinct 5 card hands can be made from a deck of 52 cards with:
3 hearts and 2 spades?

only 1 queen and 3 kings  3 red cards and 2 black cards? 
SolutionWe must use the fundamental counting principle and combinations:_{13}C_{3} × _{13}C_{2} = 286 × 78 = 22308 There are 22 308 different fivecard hands. 
SolutionWe must use the fundamental counting principle and combinations:_{4}C_{1} × _{4}C_{3} × _{48}C_{1}= 4 × 4 × 48 = 768 There are 768 different fivecard hands containing 1 queen and 3 kings 
SolutionThere are 52 ÷ 2 = 26 red cards in a deck of cards. There are 52 ÷ 2 = 26 black cards in a deck of cards. We must use the fundamental counting principle and combinations: _{26}C_{3} × _{26}C_{2} = × _{} = There are different fivecard hands containing 3 red and 2 black cards. 
How many distinct 5 card hands can be made from a deck of 52 cards with:
no fives?

4 aces?  only 3 face cards? 
SolutionWe must use the fundamental counting principle and combinations:_{4}C_{0} × _{}_{48}C_{5} = 1 × _{}1712304 = 1712304 There 1 712 304 are different fivecard hands with no fives. 
SolutionWe must use the fundamental counting principle and combinations:_{4}C_{4} × _{}_{48}C_{1} = 1 × _{}48 = 48 There are 48 different fivecard hands with 4 aces. 
SolutionWe must use the fundamental counting principle and combinations: There are 4(3) = 12 face cards in a deck of cards. _{12}C_{3} × _{}_{40}C_{2} = 220 × _{}780 = 171600 There are 171 600 different fivecard hands containing only 3 face cards. 
Parts of this work has been adapted from a Math 30 Pure learning resource originally produced and owned by Alberta Education