Name: Algebra 2
Date:
Counting (Day 2): Combinations
Objective: You will solve counting problems using combination numbers.
From the last lesson: permutations
A permutation problem is a problem that involves counting the number of ways that some of a set of things can be selected with an order. To count the number of ways that k out of n things can be selected with an order, multiply k whole numbers starting from n and counting downward. The answer to this problem is called “n permutation k” and abbreviated nPk .
New in this lesson: combinations
A combination problem is a problem that involves counting the number of ways that some things can be selected without choosing an order. Here’s an example:
Example: There are 15 tracks on a CD. You are asked to choose your 5 favorite tracks, but you don’t have to specify an order. How many ways can this be done?
To get the answer: Calculate . The answer is 3,003.
Another example:
Example: A club with 25 members wishes to select a 4-member leadership team. How many ways can this be done?
To get the answer: Calculate . The answer is 12,650.
Here is a general statement of the combination counting method:
To count the number of ways that k out of n things can be selected without an order, set up a fraction in the following way: on the top, multiply k whole numbers counting downward from n; on the bottom, multiply k whole numbers counting upward from 1. The answer to this problem is called “n combination k” and abbreviated nCk .
Note the symbol that is used to denote the answer to a combination problem. The answers to the example problems above are called 15C5 and 25C4.
How are permutations and combinations different?
Permutation problems and combination problems both involve counting the number of ways to pick a portion of the items in a set of things. But there’s an important difference between them:
· A permutation problem involves choosing with an order. That is, the same items selected in a different order would be considered to be a different choice.
· A combination problem involves choosing without an order. That is, the items are being selected as a group, and listing the items in a different order would still be considered the same choice.
Sometimes it’s necessary to read a problem carefully to recognize this distinction and make the correct choice to use a permutation number or a combination number.
You try it: Combinations
1. Find the values of these combination number symbols.
a. 5C3
b. 50C2
Directions: For each counting problem, first identify the answer as a combination number, thenwrite the answer as a fraction, and finally find the answer as a plain number.
[Example of an answer: 15C5 = = 3003.]
2. From a box of 16 color crayons, a child selects a set of 4 crayons as her favorites.
How many different ways can she make this selection?
3. There are 35 students on a football team. They must elect 3 students to serve as captains. How many different ways could the election turn out?
4. A store has to hire two cashiers. Five people are interviewed for the jobs. How many different ways can the hiring decisions be made?
5. Suppose that your English teacher asks you to choose 2 of these 4 books to read.
Beloved
Hamlet
Joy Luck Club
To Kill a Mockingbird
How many different ways could you choose?
Combination operation on your calculator
Your calculator has a shortcut operation that gives the answers to combination problems without having to multiply and divide.
For example, the answer to problem 38 is called 5C2. Here is how to find 5C2 on your calculator.
Combinations / To calculate:5C2
(“5 combination 2”) / Keys to press:
5
[MATH]3
2 [ENTER] / What you’ll see on the screen:
You try it: Combination operation on your calculator
Directions: Find the answers to these counting problems using the calculator’s combination operation as shown at the top of this page. First identify the combination needed (example: 5C2) then write the answer (example: 10).
6. From a box of 16 color crayons, a child selects a set of 4 crayons as her favorites.
How many different ways can she make this selection?
7. You go to the video store to rent some movies for the weekend. There are 1000 movies available, and you want to select 3 of them. How many ways can you make your choices?
Something different happens in this problem:
8. An essay contest is announced, with an award to be given to each of five winning entries.
Suppose that only five essays are submitted to the contest. How many ways can the awards be given?
Permutation and combination problems together
When you encounter a problem that asks you to count the number of ways that a group can be selected from a larger group, you need to figure out whether it is a permutation problem or a combination problem. Here’s a reminder of how to decide.
· If the problem situation involves the selections being made in a particular order or forspecific positions, it’s a permutation counting problem (nPk).
· If the order in which the selections are made doesn’t matter in the problem situation,
it’s a combination counting problem (nCk).
Think carefully about this distinction as you do the following problems.
You try it: Permutations and Combinations
9. From a class with 20 students, I need to select a group of 9 students to perform a top-secret mission. How many different ways can this group be selected?
a. Which of these is the correct calculation: 20P9 or 20C9 ?
b. Now use your calculator to get the answer.
10. From a class of with 20 students, I need to select a group of 9 students to play softball. Thismeans that besides choosing the students, I need to make a batting order (decide whowill bat 1st, 2nd, 3rd, etc.). How many different ways can this be done?
a. Which of these is the correct calculation: 20P9 or 20C9 ?
b. Now use your calculator to get the answer.
11. A hiker would like to invite 7 friends to go on a trip, but has room for only 4 of them.
In how many different ways can they be chosen?
12. The hiker with 7 friends wants to make a list of her 1st, 2nd, 3rd, and 4th best friends.
How many different possible lists are there?
13. An ice cream stand carries 5 flavors: vanilla, chocolate, strawberry, orange, and pineapple. The stand sells two-scoop ice cream cones.
Some examples are shown in the picture. The two flavors chosen cannot be the same. Note that vanilla-chocolate and chocolate-vanilla are considered to be different cones. In other words, the order of the scoops matters.
a. How many different kinds of twoscoop ice cream cones are there?
b. List all of the possible two-scoop ice cream cones. You can abbreviate just using the first letter of each flavor.
14. The same ice cream stand also sells two-scoop ice cream dishes. Some examples are shown in the picture. The two flavors chosen cannot be the same. In a dish, vanilla-chocolate would be considered the same dish as chocolate-vanilla, soyou shouldn’t list them both.
a. How many different kinds of twoscoop ice cream dishes are there?
b. List all of the possible two-scoop ice cream dishes. You can abbreviate just using the first letter of each flavor.
15. The answer to the ice cream dish problem (13a) was half of the answer to the ice cream cone problem (14a). Explain why this happened.
16. The ice cream stand has made some changes. There are now 7 flavors instead of 5, and it is additionally selling 3-scoop dishes and cones.
a. With 7 flavors, how many 2-scoop dishes are there?
b. With 7 flavors, how many 2-scoop cones are there?
c. With 7 flavors, how many 3-scoop dishes are there?
d. With 7 flavors, how many 3-scoop cones are there?
Thinkof your own new counting situations for the next two questions; don’t just borrow situations fromproblems you’ve already seen.
17. Make up your own problem where the answer would be found using 12P5.
18. Make up your own problem where the answer would be found using 12C5.
Recap: four counting methods
Here’s a summary of the four counting methods we’ve studied so far, along with some advice about which method to use when solving a counting problem.
· Multiplication Principle: When you have two or more decisions to make, the total number of possible ways of choosing is found by multiplying the numbers of options for each choice.
· Factorial (number of ways to put in order): If you have n objects that need to be arranged in anorder, the number of possible orders is “n factorial,” abbreviated n! .
□ You can find n! by multiplying all the whole numbers from n down to 1.
□ You can find factorials on your calculator (MATH4).
· Permutations (number of ways to select with an order): If you are asked to count the number of ways that koutof n objects can be selected with an order, the number of ways is“n permutation k,” abbreviated nPk .
□ You can find nPk by multiplying k whole numbers starting from n and counting downward.
□ You can find permutation numbers on your calculator (MATH2).
· Combinations (number of ways to select without an order): If you are asked to count the number of ways that koutof n objects can be selected without an order, the number of ways is “ncombination k,” abbreviated nCk .
□ You can find nCk by setting up a fraction in the following way:
on the top, multiply k whole numbers counting downward from n;
on the bottom, multiply k whole numbers counting upward from 1.
□ You can find combination numbers on your calculator (MATH3).
Choosing which counting method to use
· If the problem involves making two or more separate decisions and asks about the combined number of possibilities, use the Multiplication Principle.
· If the problem involves a group of objects, and asks a question about how many ways they can be selected or assigned or put in order, use the flowchart below.
Practice: mixed counting problems
19. a. Suppose that there are 3 roads connecting Town A to Town B, 4 roads connecting TownB toTown C, and 2 roads connecting Town C to Town D. Here is a picture representing the situation.
How many different ways are there to travel from Town A to Town D? (Assume that the trip goes through each of the towns just once.)
b. Make up a similar problem but involving towns W, X, Y, and Z. The answer to theproblem must be “There are 120 different ways to travel from town W to town Z.”
20. A restaurant asks to you fill out a customer satisfaction survey. It looks like this:
quality of food / excellent / good / fair / poortaste of food / excellent / good / fair / poor
variety of food / excellent / good / fair / poor
promptness of service / excellent / good / fair / poor
friendliness of service / excellent / good / fair / poor
restaurant decor / excellent / good / fair / poor
wait time for a table / excellent / good / fair / poor
Suppose that you circle one response (excellent/good/fair/poor) for each item.
How many different ways are there to fill out the survey?
21. A survey about television watching preferences asks respondents to rate each of the five major television networks (ABC, CBS, CW, Fox, NBC) on a 1-to-10 scale. Which of these is the number of different ways the survey could be completed: 510 or 105 ? Decide which, then find the value.
Directions for problems 56 through 58: Answer these counting questions. Do not use the shortcut operations under the MATH key. It’s OK to use your calculator just for arithmetic. Here’s an example of the amount of work you must show: = 35.
22. A club has 20 members.
a. In how many different ways could the club choose a president and a vice-president?
b. In how many different ways could the club choose two co-presidents?
c. In how many different ways could the club choose a president, vice-president, and secretary?
d. In how many different ways could the club choose a three-member governing committee?
e. In how many different ways could the club choose two co-presidents and a secretary?
23. a. Suppose there were 15 business people at a meeting. At the end of the meeting, each person at the meeting shook hands with every other person. How many handshakes were there?
b. If there were n people at a meeting, each shaking hands with everyone else, how many handshakes would there be?
24. Suppose that you have 12 books that are all different from each other.
a. If you want to arrange all the books in a row on a shelf, in how many different orders could the books be placed?
b. You’ve decided to let a friend borrow 3 of your 12 books. How many different ways could your friend decide which books to borrow?
Check your answers
1. a. 10 b. 1225 2. 16C4 = 1820
3. 35C3 = 6545 4. 5C2 = 10
5. 4C2 = 6 6. 16C4 = 1820
7. 1000C3 = 166,167,000 8. 5C5 = 1
9. 20C9 = 167,960 10. 20P9 ≈ 60,949,324,800
11. 7C4 = 35 12. 7P4 = 840
13. 5P2 = 20; list should include things like vc and cv but not vv
14. 5C2 = 10; list should include only one of vc and cv, not both; also not vv
15. For each pair of items on the list for 13 (like vc and cv) there’s just one (either vc or cv) for 14.
16. a. 7C2 = 21 b. 7P2 = 42 c. 7C3 = 35 d. 7P3 = 210
17–18. Answers vary, but make sure your problem for 17 involved an ordered choice and for problem 18 involved a choice without order.
19 a. 3·4·2 = 24 b. could use any three numbers whose product is 120
20. 47 = 16384
21. 105 = 100,000
22. a. 20P2 = 380 b. 20C2 = 190 c. 20P3 = 6840 d. 20C3 = 1140 e. 190·18 = 3420
23. a. 15C2 = 105 b. nC2 = n(n–1)/2
24. a. 12! = 479,001,600 b. 12C3 = 220