TOPIC 2 – COMBINATIONS MATH 30-1 CLASS BOOKLET V

Let’s go (way) back to the restaurant breakfast special for this.

Suppose the special also includes any two drinks from the following options:

1. Complete the following list every possible option for drinks.

2. Does the order matter here? (Is selecting “water, tea” and different than “tea, water”?)

3. How many selections of two drinks are there, from five possible drinks?

4. Consider the problem – how many arrangements of two objects are there, from five possible objects?

Find the answer using permutations, ?

How does that result compare to your answer from #3?

5. Put the expression 5 2 in your calculator and evaluate. How does the value of compare to that of

?

Alberta Ed Learning Outcome: Determine the number of combinations of n different elements

taken r at a time to solve problems. [C, PS, R, V]

Explore 1

coffee tea

bottled

water orange

juice fruit

smoothie

coffee, teacoffee, teacoffee, teacoffee, tea

coffee, coffee, coffee, coffee, waterwaterwaterwater

� Scenario 1: A teacher needs to choose a class president, class vice-president, and class treasurer from a group

of four students – Ava, Barack, Celine, and Dong. The following list shows all of her options.

� Scenario 2: Suppose instead the teacher needs to select any three student council members from the same

group of four. Below is the same list showing all possible arrangements of three students from the four possible.

As we’ve seen before, scenario 1 is an example of permutations, as order matters / we are arranging students.

Scenario 2, on the other hand, is an example of ________________, since ______________________________.

Refer back to the previous explore – scenario 2. Suppose the group of three students had to include Ava. How

many distinct groups would be possible now? (List them – and calculate using the combinations formula)

Explore 2

Explore 3

A,B,C

A,C,B

A,B,D

A,D,B

A,C.D

A,D.C

B,A,C

B,C,A

B,C,D

B,D,C

B,A,D

B,D,A

C,A,B

C,B,A

C,A,D

C,D,A

C,B,D

C,D,B

D,A,B

D,B,A

D,B,C

D,C,B

D,A,C

D,C,A

1. Does order matter here? How can you determine the

number of ways can she do this using the FCP?

2. …Using ?

A,B,C

A,C,B

A,B,D

A,D,B

A,C.D

A,D.C

B,A,C

B,C,A

B,C,D

B,D,C

B,A,D

B,D,A

C,A,B

C,B,A

C,A,D

C,D,A

C,B,D

C,D,B

D,A,B

D,B,A

D,B,C

D,C,B

D,A,C

D,C,A

3. Does order matter here? Cross out any redundant groups

of three from the re-printed table. How many unique

groups of three students are there?

4. How does your answer to question 3 compare to your

previous answer / when order mattered?

� Permutations involve arrangements of objects (letters, numbers, fruits, people…) – where order matters.

For example, the group of numbers 5, 10, 15 and 10, 15, 5 represent two different permutations.

� Combinations involve selections of objects where order does not matter.

So now 5, 10, 15 and 5, 10, 15 represents the same combination of numbers, written twice!

Example: Given a group of 5 people, Al, Bob, Christy, Dom, and Eve the number of ways we can:

Connect

Key Concept

The number of combinations of objects, taken at a time is less than the number

of permutations.

� If a perms and combs problem have the same and , then nPr > nCr

� Given the same ’s and ’s, the number of combs is the number of perms, divided by

the redundancy. (That is, ways any particular comb can be arranged)

That is: nCr

- Arrange any 2 of them into two desks is:

5P2

- Select any two of them to pass out tests is:

5C2

Number of COMBS of

two people

Is number of PERMS of two

people, divided by ( )

A, B

A, C

A, D

A, E

B, A

B, C

B, D

B, E

C, A

C, B

C, D

C, E

D, A

D, B

D, C

D, E

E, A

E, B

E, C

E, D

2-person PERMS, from 5 people

A, B

A, C

A, D

A, E

B, A

B, C

B, D

B, E

C, A

C, B

C, D

C, E

D, A

D, B

D, C

D, E

E, A

E, B

E, C

E, D

2-person COMBS, from 5 people

10 unique combinations.

Order doesn’t matter

Greyed-outs are redundant

when countin’ combs.

20 unique permutations

Since each of the 10 combinations can be

arranged ways

Number of combinations = Number of permutations

Redundancy

, number of ways each comb can be arranged

On formula sheet:

Many combinations problems involve considering multiple cases.

For example, suppose a teacher with a class of 30 ambitious math students announces that he’ll bring them all pie

if at least 20 of them get a B or better on their Combinatorics test.

We can break this down into cases. AT LEAST 20 students (out of 30) means: 20 or 21 or 22 or 23 or …. or 29 or 30

Perms v Combs!

Something you probably use every day is a combination lock.

Suppose after a particularly late night of – studying, you forgot your “combination”.

But you were pretty sure the 3-number combo used some variation of the numbers

5, 10, 15, or maybe 20. What is the maximum number of attempts you’d need to crack

that baby open?

“AT LEAST” type problems

There are 11 specific cases here!

Know your formula sheet!

Key Concept

For “At least” type problems,

break down into specific cases

1. Decide whether each of the following is a PERMS or COMBS question. Then – solve it!

(a) In a volleyball league with 7 teams, every team must play the other once. How many games would need

to be scheduled?

(b) A circle has five points marked on it. How many unique triangles can be formed?

(c) In a class of 15 students, a teacher needs four volunteers to move a table. How many ways can she select

the four students?

(d) A teacher needs four volunteers for a class demo. One student will ask her fellow students questions, one

will record the results, one will put the results on a chart, and the final student will report on the results

back to the class. How many unique ways can the teacher do this, if the class has 15 students?

(e) From a committee of 9 people, a 3-person sub-committee must be formed. How many ways can this be

done?

(f) From a committee of 9 people, a sub-committee must be formed consisting of a chairperson, vice-chair,

and secretary. How many ways can this be done?

(g) A standard deck of 52 cards contains 13 of each suit. (Suits are Hearts, Diamonds, Clubs, and Spades)

Given a five-card hand, how many ways can a hearts flush be dealt? (Flush means all five cards are the same

suit)

(h) How many 3-letter words are possible using the letters in the word “MATH”?

(i) A car lot has 22 different new cars for sale. How many ways can a car rental agency purchase 10 of them?

(j) A car rental agency has 11 different cars available on-site. Ricardo, Beth, and Kidist walk in, each wanting

to rent a car. How many different ways can three cars be rented to these customers?

Practice

2. Use the formula for nCr to evaluate each, then check using the nCr function on your calculator.

(a) 8C5 (b) 4C3

3. Algebraically solve each equation:

(a) (b)

4. List all possible permutations and combinations of A, B, and C taken two at a time. Demonstrate the

relationship between the number of perms and the number of combs.

5. Hugo reaches into his pocket and finds a loony, a quarter, a dime, a nickel, and a penny. How many different

sums of money could he make?

6. A pizza shop has 15 choices of toppings. They are having a special where any four-topping pizza is available for

$14.99. How many options does a customer have, if one of the toppings must be anchovies?

7. The girls in a small math class consist of Melanie, Sandeep, Bianca, Amy, Maria, Chandra, and Janice.

(a) How many groups can be formed consisting of exactly 4 females?

(b) How many smaller groups of 4 can be formed if Melanie must be in the group?

(c) How many smaller groups of 4 can be formed if Amy and Maria can’t both be in any one group? (They’re

fighting)

(d) How many groups can be formed that consist of at least four females?

8. A class consists of 8 boys and 10 girls. How many ways can a class student council be formed if:

(a) There must be exactly 2 boys and 2 girls in the council.

(b) There must be exactly 2 boys / 2 girls – and a particularly brainy student Jefferson must be on the council.

(c) (For this one – forget the 2 boys / 2 girls requirement. Start fresh!) There must be at least 1 girl on the

council.

9. From a standard 52-card deck, five cards are randomly selected. How many distinct hands are possible?

10. Refer back to the previous question.

(a) How many five-card hands would consist of 3 Kings and 2 Queens?

(b) How many five-card hands would consist of a flush?

(c) How many five-card hands would include 4 aces?

(d) How many five-card hands would have at least two deuces? (two deuces, or three deuces, and so on)

Diploma

Example

11.

Mixed Questions (Could be perms or combs)

Diploma

Example

12.

`

Diploma

Example

13.

SE How many different 4-letter arrangements are possible using any 2 letters from the

word SMILE and any 2 letters from the word FROG?

Diploma

Example

14.

Diploma

Example

15. In a group of 9 people, there are four females and five males. Determine the number of four-

member committees consisting of at least one female that can be formed.

Diploma

Example

16. In a group of 9 people, there are four females and five males. A four-member committee must

be chosen that consists of a chairperson, a vice-chair, and two regular members. How many

such committees can be chosen?