Wize High School Grade 12 Pre-Calculus Textbook > Permutations and Combinations [BC/AB]

Combinations

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Combinations 
Combinations are different from permutations -- order does NOT matter!

Example
Find the number of ways you can draw 4 gumballs out of a bag of 10 different coloured gumballs, one at a time →    permutation question   
Find the number of ways you can draw 4 gumballs out of a bag of 10 different coloured gumballs all at once →    combination question   
Combination Formula
The number of ways to choose k out of n different objects is
(nk)=n!k!(n−k)!{\color{red}n\choose \color{orange}k}=\frac{\color{red}n\color{black}!}{\color{orange}k\color{black}!(\color{red}n\color{black}-\textcolor{orange}{k})!}(kn​)=k!(n−k)!n!​
This notation is read as "n choose k".

Properties 
(n0)=1{n\choose 0}=1(0n​)=1
(nn)=1{n\choose n}=1(nn​)=1
(n1)=n{n\choose 1}=n(1n​)=n
(nk)=(nn−k)\binom{n}{k}=\binom{n}{n-k}(kn​)=(n−kn​)

Wize Tip
Use some variation of this combination formula if the question involves keywords "choose", "pick", "select" with no order.

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Example: Combination Notation
Evaluate the expression (108) (81) (77)\binom{10}{8}\ \binom{8}{1}\ \binom{7}{7}(810​) (18​) (77​).

Let's first simplify using some of the properties:
=(108)(8)(1)=\binom{10}{8}\left(8\right)\left(1\right)=(810​)(8)(1)

Now we simplify the first bracket using the definition of "n choose k"
=10!8!(10−8)!(8)(1)=\frac{10!}{8!\left(10-8\right)!}\left(8\right)\left(1\right)=8!(10−8)!10!​(8)(1)
=10!8!2!(8)=\frac{10!}{8!2!}\left(8\right)=8!2!10!​(8)
=10!(8×7!) 2!(8)=\frac{10!}{\left(8\times7!\right)\ 2!}\left(8\right)=(8×7!) 2!10!​(8)
=10!7! 2!=\frac{10!}{7!\ 2!}=7! 2!10!​
=10×9×8×7!7! 2!=\frac{10\times9\times8\times7!}{7!\ 2!}=7! 2!10×9×8×7!​
=10×9×82!=\frac{10\times9\times8}{2!}=2!10×9×8​
=10×9×82=\frac{10\times9\times8}{2}=210×9×8​
=10×9×4=10\times9\times4=10×9×4
=360=360=360

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There's a certain type of questions called "labelling problems" that involve assigning members in a group to different labels (categories)

Example: Splitting up a Group w/ Combination
In how many ways can you split up a group of 14 students so 3 of them go to science class, 4 of them go to math class, and the remaining students go to English class?

Since the order in which we assign there students class doesn't matter, it's a combination question.
Science class: (143)\binom{14}{3}(314​)
Math class: since there are now only 11 students left to choose from--(114)\binom{11}{4}(411​)
English class: the remaining 7 students have to go to English -- (77)\binom{7}{7}(77​)
Therefore, there are (143) (114) (77)=14!3! 11!×11!4!7!×7!7!0!=14!3!4!7!\binom{14}{3}\ \binom{11}{4}\ \binom{7}{7}=\frac{14!}{3!\ \color{blue}11!}\times\frac{\color{blue}11!}{4!\color{red}7!}\times\frac{\color{red}7!}{7!0!}=\frac{14!}{3! 4! 7!}(314​) (411​) (77​)=3! 11!14!​×4!7!11!​×7!0!7!​=3!4!7!14!​

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Example: Combinations
A freshman class has 15 students, including Adam, Ali, Ace, and Allan. How many committees of four can be formed from this class if 
a) all members of the group will take on equal roles?
b) the 4 committe members will take on the roles of representatives for grades 9-12 respectively?
c) exactly two of the members must be Adam, Ali, Ace, or Allan?

PART a)
If all 4 members will take on the same roles, the order in which they are chosen doesn't matter.
Therefore, there are (154)\binom{15}{4}(415​) such ways of forming such committees.

PART b)
Since the 4 roles are distinguishable, we can think of this problem as picking the committee members one at a time and assigning them the roles of representative for grades 9-12 in order. So this is like say "in how many ways can you assign 4 different roles from 15 students?"
Therefore, there are 15×14×13×12  or 15!11!15\times14\times13\times12\ \ \text{or}\ \frac{15!}{11!}15×14×13×12  or 11!15!​ such ways of forming this committee

PART c)
Let's choose our 2 members from the specified subgroup first: (42)\binom{4}{2}(24​)
We now have to pick 2 more members from the remaining 11 students: (112)\binom{11}{2}(211​)
Therefore, there are (42) (112)\binom{4}{2}\ \binom{11}{2}(24​) (211​) ways to form this committee

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Example: Combinations w/ Cards
In how many ways can you draw three cards from a standard deck if:

a) all three cards are from the same suit?

There are 4 cases here -- all cards of suit type 1 (diamonds), suit type 2 (hearts), suit type 3 (clubs) or suit type 4 (spades). Each case has the same number of outcomes.

For each suit, we need to pick 3 of the 13 cards: (133)\binom{13}{3}(313​)

Therefore, there are 4×(133)4\times\binom{13}{3}4×(313​) ways to draw 3 cards like this.

b) at least one of the cards is a Jack?

Let's use the complement.

No restrictions: choose 3 out of 52 cards  →  (523)\binom{52}{3}(352​)

Complement: no jacks (pick 3 out of the remaining 48 cards)  → (483)\binom{48}{3}(348​)

Therefore, there are (523)−(483)\binom{52}{3}-\binom{48}{3}(352​)−(348​) ways to draw 3 cards like this.

Practice: Combination Notation
Which of the following is false?

a. (123)=(129)\binom{12}{3}=\binom{12}{9}(312​)=(912​)

b. (120)=1\binom{12}{0}=1(012​)=1

c. (1212)=12\binom{12}{12}=12(1212​)=12

d. (123) (95) (43)=12!3! 5! 3!\binom{12}{3}\ \binom{9}{5}\ \binom{4}{3}=\frac{12!}{3!\ 5!\ 3!}(312​) (59​) (34​)=3! 5! 3!12!​

e. All of these statements are true

I don't know

Practice: Combinations
A committee of 7 is formed from a group of 10 men and 9 women. In how many ways can the committee be formed if 

a) at least one of the members is male?
b) at least one of the member is male and at least one of the members is female?

a) at least one of the members is male?

a.(101)(186){10\choose 1}{18\choose 6}(110​)(618​)

b.(101)(96){10\choose 1}{9\choose 6}(110​)(69​)

c.(197)−(97){19\choose 7}-{9\choose 7}(719​)−(79​) 

d.(197){19\choose 7}(719​)

I don't know

Practice: Combinations w/ Cards
A full-house is a 5-card hand that consists of 3 cards of the same value and a pair. For example, 3 kings and 2 fives. 

In how many ways can you draw a full-house?

a. (523) (492)\binom{52}{3}\ \binom{49}{2}(352​) (249​)

b. (131) (121) (43) (42)\binom{13}{1}\ \binom{12}{1}\ \binom{4}{3}\ \binom{4}{2}(113​) (112​) (34​) (24​)

c. (131) (43) +(131) (42)\binom{13}{1}\ \binom{4}{3}\ +\binom{13}{1}\ \binom{4}{2}(113​) (34​) +(113​) (24​)

d. (523) (43)+(493) (42)\binom{52}{3}\ \binom{4}{3}+\binom{49}{3}\ \binom{4}{2}(352​) (34​)+(349​) (24​)

e. (131) (43)+(121) (42)\binom{13}{1}\ \binom{4}{3}+\binom{12}{1}\ \binom{4}{2}(113​) (34​)+(112​) (24​)

I don't know

Practice: Combinations
Josh is taking a 10 question test. In how many ways can he get a score of 8 out of 10 if each question on the test is

i) a True/False question, where each answer is either "True" or "False"?
ii) a 5-option multiple choice question, where only 1 of the 5 options is correct?

i) a True/False question, where each answer is either "True" or "False"?

a. (102)\binom{10}{2}(210​)

b. (102)×510\binom{10}{2}\times5^{10}(210​)×510

c. (102)×42\binom{10}{2}\times4^2(210​)×42

d. (102)×210\binom{10}{2}\times2^{10}(210​)×210

e. 2102^{10}210

I don't know