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RTD’s Math 30-1
December 2014
PERMS and COMBS WORKSHOP
www.rockthediploma.com
Solutions are on www.rtdmath.com
Develop algebraic and numeric reasoning that involves combinatorics.
• Students are expected to simplify expressions involving 𝑛𝑃𝑟, 𝑛𝐶𝑟, and 𝑛! algebraically.
• Probability is beyond the scope of this course.
1) The FUNDAMENTAL COUNTING PRINCIPAL (Arrangements 1)
Apply the fundamental counting principle to solve problems.
• Alternative methods of problem solving involving pictorial or visual explanations (tree diagrams, lists) are
appropriate.
 Apply the fundamental counting principle to various problems involving at most two cases or constraints
SE Apply the fundamental counting principle to various problems involving three or more cases or constraints
 Example 1: A continental breakfast comes with a choice of one hot beverage (coffee or tea), one cold beverage
(orange, apple, or grapefruit juice), one fruit (apple, banana, grapes, or melon) and one pastry (doughnut, muffin,
croissant, turnover, or strudel)
Concept – Fundamental Counting Principal
 Example 2: A child’s doll comes with a choice of three hair colours (blonde, brown, or black) and
 Example 3: How many four digit numbers are possible, if the last digit must be odd?
 Example 4: How many license plates are possible in a jurisdiction where each plate must begin with 3 letters that
don’t repeat (and the first letter must be a consonant) – followed by any 4 digits. (Repetition allowed)
 Example 5: The final score of a hockey game was Jets 3, Oilers 2. How many scores were possible at the end of the
third period?
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2) PERMUTATIONS (Arrangements 2)
Determine the number of permutations of 𝒏 elements taken 𝒓 at a time to solve problems.
• Permutation problems could involve repetition of like elements and constraints.
• Circular and ring permutations are beyond the scope of this outcome.
• Single 2-dimensional pathways may be used as an application of repetition of like elements.
 Recognize and address problems involving the terms and or or
SE Recognize and address problems involving the terms at least or at most, and scenarios involving cases where items cannot be
together
SE solve problems involving permutations when two or more elements are identical (repetitions), with more than one constraint
Three students are running in a 200m dash race.
Ti
Chris
1. Complete the list below to show all the possible ways they can
finish first, second and third. (How many possibilities are there?
One possible order is listed to get you started)
Rad
Chris, Ti, Rad
tea
Chris, Rad, Ti
tea
2. Write a number for each blank that provides an expression that shows how to find the answer to #1 using
the fundamental counting principal.
Jon
Number of possibilities
for finishing first
…finishing second
…finishing last
click
Then scroll
to “PRB”
3. Enter the following expression in your calculator: 3! (Read: “3 factorial”). Make a conjecture about how 3! Is
calculated. How does your answer for the previous question relate to the value of 3! ?
4. The top finisher qualifies for school finals, while the second place finisher gets to enter consolation race.
Complete the list of all the possible top two finishes. (How many possibilities are there?)
Chris, Ti
first
second
5. Enter the expression “3 𝑛𝑃𝑟 2” and “3 𝑛𝑃𝑟 3” in your calculator. (Found in the same place…”math”…”prb”) How does the
value you obtain compare to your answers for the previous questions?
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Finally, suppose this time there are four students running in a race. We firstly want to consider all of possible outcomes
of this race, in terms of how the students finish. (From 1st to 4th place)
Don’t feel like writing out all possible outcomes of a four-person race? Don’t worry, I’ve got your back. But I will abbreviate each
student by the first letter of their name….
Abi
Bryn
Cole
List of all possible arrangements
of all 4 racers:
Dom
A,B,C,D
B,A,C,D
C,A,B,D
D,A,B,C
A,B,D,C
B,A,D,C
C,A,D,B
D,A,C,B
A,C,B,D
B,C,A,D
C,B,A,D
D,B,A,C
A,C,D,B
B,C,D,A
C,B,D,A
D,B,C,A
A,D,B,C
B,D,A,C
C,D,A,B
D,C,A,B
A,D,C,B
B,D,C,A
C,D,B,A
D,C,B,A
1. How many possible arrangements of the four students are possible? How can this answer also be obtained using
the fundamental counting principal?
2. Suppose we are giving 1st, 2nd, and 3rd place ribbons
out. How many possible arrangement are there of
the top three finishers?
See the table on the right, identical to the original table but with
the fourth student (now irrelevant) greyed-out.
Does the resulting list cover all “Top 3” possibilities?
List of all possible arrangements
of the TOP 3 racers:
A,B,C,D
B,A,C,D
C,A,B,D
D,A,B,C
A,B,D,C
B,A,D,C
C,A,D,B
D,A,C,B
A,C,B,D
B,C,A,D
C,B,A,D
D,B,A,C
A,C,D,B
B,C,D,A
C,B,D,A
D,B,C,A
A,D,B,C
B,D,A,C
C,D,A,B
D,C,A,B
A,D,C,B
B,D,C,A
C,D,B,A
D,C,B,A
3. Next we’ll consider the number of possible arrangements of top-two finishers. (Winner & runner-up) This time, we
grey-out each “3rd” letter as well. Does the remaining list cover all remaining “top 2” possibilities?
(Or are there repetitions? Ok I’ll give it a away. YES, as you see here,
the first two 2-student arrangements are identical. So cross out the
second one. Go ahead, scratch a line right through it. Don’t need it,
it’s already covered by the first “AB” right above it. Then, keep it
going. Cross out each redundant two-letter arrangement.
 How many top-two finishers (2 student / letter arrangements) are
possible?
 How many 1-student arrangements are there? (How many ways
are there for a student to finish first?)
List of all possible arrangements
of the TOP 2 racers:
A,B,C,D
B,A,C,D
C,A,B,D
D,A,B,C
A,B,D,C
B,A,D,C
C,A,D,B
D,A,C,B
A,C,B,D
B,C,A,D
C,B,A,D
D,B,A,C
A,C,D,B
B,C,D,A
C,B,D,A
D,B,C,A
A,D,B,C
B,D,A,C
C,D,A,B
D,C,A,B
A,D,C,B
B,D,C,A
C,D,B,A
D,C,B,A
4. Connecting this all together.
 The number of way to arrange 3 “things” out of a group of 4, is the same as the number of
ways to arrange all four things.
 The number of way to arrange 2 “things” out of a group of 4, is the number of ways to arrange
all four things, divided by 2. (More specifically – divided by the number of ways to arrange 2 things)
 So, the number of way to arrange 1 “thing” out of a group of 4, is the number of ways to
arrange all four things, divided by the number of ways to arrange the 3 things left over.
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 Example 1: How many ways can all of the letters in the word “EXAM” be arranged?
Concept – Factorial Notation
 Example 2: A small class of 10 students (5 boys, 5 girls) are to be arranged in a single line.
(a) How many ways can this be done?
(b) … Suppose we had to have all of the girls first, then the boys?
(c) … Suppose the girls have to be all together?
Concept – Arrangements were certain elements must be kept together
Count the objects that must
be kept together as one unit
Number of permutations, counting
the “together objects” as one
×
Number of permutations of the
“together objects” among
themselves
…Solution!
 Example 3: How many “words” can be made from all of the letters in “turnpike” if all of the vowels must be kept
together?
 Example 4: A family consisting of mom, dad, 2 boys, and 3 girls line up for a portrait. How many ways can the family
be arranged if all of the children must be kept together?
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Concept – Permutations of 𝒏 objects taken 𝒓 at a time
 Example 5a: How many distinct four-letter arrangements are possible using letters in the word “chemistry”.
 Example 5b: How many distinct four-letter arrangements are possible using letters in the word “chemistry”, if the
first letter must be a “c”.
 Example 5c: How many distinct arrangements are possible using all of the letters in the word “chemistry”, if the first
eight letters must be “chemistr”.
 Example 5d: How many distinct arrangements are possible using all of the letters in the word “chemistry”, if each
“word” must begin with all of the consonants. (c, h, m, s, t, r)
 Example 5e: How many distinct arrangements are possible using all of the letters in the word “chemistry”, if all of the
consonants must be kept together?
 Example 6: For a play with 3 male and 4 female parts, a drama teacher has 5 boys and 7 girls try out. How many
distinct casts are possible?
 Example 7: How many distinct arrangements are possible using all of the letters in the word “bee”?
 Example 8: How many distinct arrangements are possible using all of the letters in the word “beee”?
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Permutations of objects with repetitions
A grocery store manager is making a fruit arrangement in a display window. (Could happen) He has
an apple, an orange, and two banans – four items of fruit in total.
How many arrangements are possible?
Would the display be different if he changed the position of the last two fruits?
1. Below is a list of all possible arrangements of the letters “A”, “O”, “B1 ”, “B2 ”. Cross out any
redundant (repeated) entries – that is, entries that represent an identical fruit arrangement.
𝐴, 𝑂, 𝐵1 ,𝐵2
𝐵1 , 𝑂, 𝐴, 𝐵2
𝐵2 , 𝐴, 𝐵1 , 𝑂
𝑂, 𝐴, 𝐵1 , 𝐵2
𝐴, 𝑂, 𝐵2 ,𝐵1
𝐵1 , 𝐴, 𝑂, 𝐵2
𝐵2 , 𝐴, 𝑂, 𝐵1
𝑂, 𝐴, 𝐵2 , 𝐵1
𝐴, 𝐵1 , 𝑂, 𝐵2
𝐵1 , 𝐵2 , 𝐴, 𝑂
𝐵2 , 𝐵1 , 𝐴, 𝑂
𝑂, 𝐵1 , 𝐴, 𝐵2
𝐴, 𝐵2 , 𝑂, 𝐵1
𝐵1 , 𝐵2 , 𝑂, 𝐴
𝐵2 , 𝑂, 𝐴, 𝐵1
𝑂, 𝐵2 , 𝐴, 𝐵1
𝐴, 𝐵1 , 𝐵2 , 𝑂
𝐵1 , 𝑂, 𝐵2 , 𝐴
𝐵2 , 𝐵1 , 𝑂, 𝐴
𝑂, 𝐵1 , 𝐵2 , 𝐴
𝐴, 𝐵2 , 𝐵1 , 𝑂
𝐵1 , 𝐴, 𝐵2 , 𝑂
𝐵2 , 𝑂, 𝐵1 , 𝐴
𝑂, 𝐵2 , 𝐵1 , 𝐴
2. How many distinct arrangements are there of four fruits, consisting of 1 apple, 1 orange, and
2 bananas?
3. How many distinct arrangements are there if the first two fruits must be bananas?
Must be bananas!
4. How many distinct arrangements are there if the first fruit must be a banana?
Must be a banana!
5. Suppose we replaced each orange with a banana. (crazy!) How many distinct arrangements
would there be then?
𝐴, 𝐵3 , 𝐵1 ,𝐵2
𝐵1 , 𝐵3 , 𝐴, 𝐵2
𝐵2 , 𝐴, 𝐵1 , 𝐵3
𝐵3 , 𝐴, 𝐵1 , 𝐵2
𝐴, 𝐵3 , 𝐵2 ,𝐵1
𝐵1 , 𝐴, 𝐵3 , 𝐵2
𝐵2 , 𝐴, 𝐵3 , 𝐵1
𝐵3 , 𝐴, 𝐵2 , 𝐵1
𝐴, 𝐵1 , 𝐵3 , 𝐵2
𝐴, 𝐵2 , 𝐵3 , 𝐵1
𝐴, 𝐵1 , 𝐵2 , 𝐵3
𝐴, 𝐵2 , 𝐵1 , 𝐵3
𝐵1 , 𝐵2 , 𝐴, 𝐵3
𝐵1 , 𝐵2 , 𝐵3 , 𝐴
𝐵1 , 𝐵3 , 𝐵2 , 𝐴
𝐵1 , 𝐴, 𝐵2 , 𝐵3
𝐵2 , 𝐵1 , 𝐴, 𝐵3
𝐵2 , 𝐵3 , 𝐴, 𝐵1
𝐵2 , 𝐵1 , 𝐵3 , 𝐴
𝐵2 , 𝐵3 , 𝐵1 , 𝐴
𝐵3 , 𝐵1 , 𝐴, 𝐵2
𝐵3 , 𝐵2 , 𝐴, 𝐵1
𝐵3 , 𝐵1 , 𝐵2 , 𝐴
𝐵3 , 𝐵2 , 𝐵1 , 𝐴
6. How does the number of distinct arrangements of 𝑛 objects relate to the number of
repetitions?
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Concept – Permutations of 𝒏 objects where some items are identical
When arranging objects - where of some of the objects are identical – we divide the repetition.
BOO
Consider the letters in the word “BOO”
BOO
There are 3! ways to arrange the letters…
OBO
OBO
OOB
OOB
But many of these arrangements are identical,
since re-arranging the O’s among themselves
does not change the word!
 There are 2! ways to arrange the O’s for any
of the 3! arrangements. This means 2! of the
3! arrangements are redundant.
“Divide them out”!
Unique
Arrangements:
𝟑!
=𝟑
𝟐!
In general, when determining the number of unique arrangements of “𝑛” objects, where there are
“𝑎” identical objects of one type, “𝑏” identical objects of another type, and so on…
The number of unique arrangements is given by:
𝒏!
𝒂! 𝒃! 𝒄! …
 Example 9: (a) How any distinct arrangements are there of the letters in the word BANANA?
(b) … if the first letter must be a “B”?
(c) … if the first letter must be an “A”?
(d) … if the As must all be kept together.
 Example 10: A car lot manager wants to line up 10 cars that are identical except for colour. There are 3 red cars, 2
blue cars, and 5 green cars. Determine the number of possible arrangments of the 10 cars if
(a) There are no restrictions
(b) They are lined up in a row along one side of a parking lot, and a blue car is parked on each end of the row
(c) The blue cars cannot be together
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3) COMBINATIONS
Determine the number of combinations of n different elements taken r at a time to solve
problems.
• Students are expected to know both 𝑛𝐶𝑟 and
𝑛
notation.
𝑟
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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)
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Concept – Permutations of 𝒏 objects taken 𝒓 at a time
The number of combinations of 𝑛 objects, taken 𝑟 at a time is ____________ the
Key Concept number of permutations.
 If a perms and combs problem have the same 𝑛 and 𝑟, then nPr ___ nCr
 Given the same 𝑛’s and 𝑟’s, the number of combs is the number of perms, divided by
On formula sheet:
the redundancy. (That is, ways any particular comb can be arranged)
That is: nCr =
⬚
𝒏𝑷𝒓
𝒓!
Example: Given a group of 5 people, Al, Bob, Christy, Dom, and Eve the number of ways we can:
-
Arrange any 2 of them into two desks is:
-
Select any two of them to pass out tests is:
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 Solve problems involving permutations or combinations
 SE Solve problems involving both permutations and combinations
 Example 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?
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 Example 2: 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.
 Example 3: A pizza shop has 15 choices of toppings. They are having a special where any four-topping pizza is
available for $14.99.
(a) How many options does a customer have?
(b) How many options are there if the pizza cannot have anchovies?
 Example 4: Refer back to the previous question. Sometimes customers choose fewer toppings – as few as none.
How many options does a customer have given that the pizza could have at most four toppings?
 Example 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?
 Example 6: 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?
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 Example 7: 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.
 Example 8: From a standard 52-card deck, five cards are randomly selected. How many distinct hands are possible?
 Example 9: 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)
 Solve for n in equations involving one occurrence of 𝑛𝑃𝑟 or 𝑛𝐶𝑟 given 𝑟, where 𝑟 ≤ 3 , and identify extraneous solutions
 Example 10: Algebraically solve each equation:
(a) 𝑛𝐶2 = 28
(b)
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 Obtain solutions to problems involving at most two cases or constraints SE … involving three or more cases or constraints
Mixed Questions (Could be perms or combs!)
 Example 11:
Diploma
Example
 Example 12:
Diploma
Example
 Example 13: Consider the word “JACKSON”. How many ways are there to…
(b) Arrange all of the letters in the word – if the
(a) Arrange all of the letters in the word
first two letters must be “JA” (in that order)
(d) Arrange all of the letters in the word – if the
letter must be a consonant
(c) Make an arrangement using any two letters.
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 Example 14: A school play has three male parts. The drama club has 7 boys.
(a) Determine the number of arrangements of boys for the three parts, using
the fundamental counting principal
(b) Determine the number of arrangements of boys, using the formula
𝑛𝑃𝑟
=
𝑛!
(𝑛 − 𝑟)!
(c) Determine the number of arrangements of boys possible if Wagner has been promised the lead role.
(d) Determine the number of arrangements if Wagner has been promised a role.
 Example 15: In a group of 9 people, there are four females and five males. Determine the number of fourmember committees consisting of at least one female that can be formed.
 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?
 Example 17: A co-ed basketball team, consisting of 4 boys and 3 girls, is lining up for a photo.
(a) How many different arrangements of the team are possible?
(b) How many arrangements are possible if all of the boys are on the left side, girls on the right side?
(c) How many arrangements are possible if the boys and girls must alternate?
(d) How many arrangements are possible if the boys all must be kept together?
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 Example 18:
Diploma
Example
 Example 19:
Diploma
Example
 Example 20: SE If all of the letters in the word DIPLOMA are used, then how many different arrangements
are possible that begin and end with an I, O, or A?
 Example 21: SE How many different 4-letter arrangements are possible using any 2 letters from the
Diploma
Example
word SMILE and any 2 letters from the word FROG?
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4) BINOMIAL EXPANSION
Expand powers of a binomial in a variety of ways, including using the binomial theorem
(restricted to exponents that are natural numbers).
• Teachers may choose to show the relationship between the rows of Pascal’s triangle and the numerical
coefficients of the terms in the expansion of a binomial (𝑥 + 𝑦)𝑛 , 𝑛 ∈ 𝑁.
• Students are expected to recognize various patterns in the binomial expansion.
 Pascal’s Triangle
1
Blaise Pascal is long deceased multi-discipline scholar who formally described the
pattern of numbers that now bears his name.
1
Have a good, critical look at the triangular array of numbers provided. Then,
 Jot down at least three patterns you notices. (Compare with a neighbor)
1
1
1
2
3
1
3
1
 Complete the next three rows of the triangle.
Needed – scrap paper, basic FOIL skills, some slightly advanced FOIL-ing skills, a little bit of heart
 Expanding a binomial in the form (𝒙 + 𝒚)𝒏
In this exploration we’ll look at a binomial – representing “any” binomial – in the form (𝑥 + 𝑦), taken to various degrees.
-
Fill out the first five rows of the following table. Do your work on scrap paper.
-
Once you get to the fourth row, things will get a bit intense! See me if you are unsure of
how to expand a binomial to degree 3. (And beyond!) Make sure you write out your
expanded forms in descending order of degree. (DOOD)
-
Once you’ve completed five rows, take stock of all of the patterns you see. At this point,
Tip: Centre your expanded expressions in the table
we’ll stop the madness, and complete the rest by patterns only.
(𝑥 + 𝑦)0
(𝑥 + 𝑦)1
(𝑥 + 𝑦)2
(𝑥 + 𝑦)3
(𝑥 + 𝑦)4
(𝑥 + 𝑦)5
(𝑥 + 𝑦)6
Lil Ol’ List of PATTERNS I can see in the expansion of (𝒙 + 𝒚)𝒏 :
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Fully expanding a binomial using Pascal’s Triangle
 Complete these four easy steps to see the expanded form of (𝟑𝒙 + 𝟐𝒚)𝟒 !
Step  - Write out a set of blank “triple brackets” for each term in the expansion. How many terms will there be if 𝑛 = 4?
(This step has been started below, please complete!)
(
)(
)(
)+(
)(
)(
)+
Step  - In each first bracket above, write the corresponding Pascal’s triangle term – representing the coefficient.
Step  - In each second bracket above, place the first term of the binomial to be expanded. In each third bracket, place the second term.
Step  - Exponents of 𝑥 (here meaning the first term) go from 𝑛 down to 0. Write each exponent of 𝑥 above.
Step  - Exponents of 𝑦 (here meaning the second term) go from 0 up to 𝑛. Write each exponent of 𝑥 above.
Step  - Almost done – simplify your expression into a nice, clean, bracketless and DOOD form.
So far so fun, but what’s this got to do with PERMS? (Or COMBS, for that matter)
 Refer back to Pascal’s triangle. Please note that the very first entry (“1”), can be obtained by a combinations
expression, namely 0𝐶0 . The rest of the entries can similarly be determined using the appropriate combs expression.
State the comb term equivalent to each triangle number (test on your calculator) – and, as always, make note of the
pattern!
1st row 
2nd row 
1
3rd row 
1
4th row 
5th row 
0𝐶0
1
1
1
2
3
4
1𝐶1
1
1
3
6
1
4
1
𝒏𝒕𝒉 row 
 Use the patterns you noticed to write out the combs terms for the 𝑛𝑡ℎ row of Pascal’s triangle. How many terms will
there be on the 𝑛𝑡ℎ row?
19
In explore 3, we looked at the expanded form of a binomial (3𝑥 + 2𝑦)4 . Next, we’ll kick it up a notch and look at the
expanded form of (𝟑𝒙 + 𝟐𝒚)𝟓 ! Our goal will be developing a method to determine any particular term, without fully
expanding the thing out.
Before we get started, let’s review some notation.
 𝑥 is the first term (Here, 𝑥 = 3𝑥)
 𝑦 is the second term (Here, 𝑦 = 2𝑦)
 𝑛 is the binomial exponent. (Here, 𝑛 = 4)
The Example Question: What is the third term in the expansion of (3𝑥 + 2𝑦)4 ?
Let’s start by looking at its fully expanded form
(3𝑥 + 2𝑦)4 expands to: 243𝑥 5 + 810𝑥 4 𝑦 + 𝟏𝟎𝟖𝟎𝒙𝟑 𝒚𝟐 + 720𝑥 2 𝑦 3 + 240𝑥𝑦 4 + 32𝑦 5
Yep, there it is. The third term. (𝑡3 ) We’ll now look
at a way to find this without fully expanding.
Developing a method to find:
The third term of (3𝑥 + 2𝑦)5 , while also developing a formula for any term in the expansion of (𝑥 + 𝑦)𝑛
Firstly, notice that if we wanted the first term, we know the
exponent of 𝑦 is 0. For the second term, it’s 1, and so on.
(The exponent of 𝑦 bumps up one each successive term)
So, designate a variable “𝑘” to be the power of 𝒚. (𝑦 being simply
nd
whatever the 2 term is in the original binomial we’re expanding)
Key Concept:
"𝒌" = 𝑒𝑥𝑝 𝑜𝑓 𝒚
YOU SAY: Find the third term, 𝑡3
We want the third term, 𝑡3 , and for it the exponent of 𝑦 is 2, so 𝑘 = 2.
I SAY: Find the “𝑘 + 1"𝑡ℎ term, 𝑡2+1
And we’re saying the same thing!
rd
THIS IS BIG: The term we want (3 ), is one higher than 𝑘
The term where the
power of 𝑦 is “2”
 We want the “𝑘 + 1"𝑡ℎ term.
Now, remember for a binomial in the form (𝑥 + 𝑦)𝑛 , we use the
“𝑛 + 1"𝑡ℎ row of Pascal’s Triangle. Here, 𝑛 = 4, so we’d use the fifth row.
1
1
1
1
1
4C0
4C1
Finally, notice that for the third term, 𝟏𝟎𝟖𝟎𝒙𝟑 𝒚𝟐 , the exponent of 𝑦 is 𝑘 (of
1
2
3
4
4C2
course), and the exponent of 𝑥 is 𝑛 − 𝑘. (𝑛 = 5 and 𝑘 = 2)
1
3
6
So we have our formula: 𝑡𝑘+1 = 𝑛𝐶𝑘 (𝑥)𝑛−𝑘 (𝑦)𝑘
1
4
And, 𝑡3 = 𝑡2+1 = 5𝐶2 (3𝑥)5−2 (2𝑦)2
1
Notice that the pascal’s-derived
coefficient is 4C2 that is, nCk
⬚
, that is, 𝒏𝑪𝒌
= 10(27𝑥 3 )(4𝑦 2 )
= 𝟏𝟎𝟖𝟎𝒙𝟑 𝟒𝟐
20
 Demonstrate an understanding of patterns that exist in the binomial expansion
 Expand (𝑥 + 𝑦)𝑛 , 𝑛 ∈ 𝑁 or determine a specified term in the expansion of a binomial with linear terms
 Expand (𝑥 + 𝑦)𝑛 , 𝑛 ∈ 𝑁 or determine a specified term in the expansion of a binomial with non-linear terms
 Determine an unknown value in (𝑥 + 𝑦)𝑛 , given a specified term in its expansion
 Example 1: For each, describe the next Pascal’s triangle row in terms of combinations. (In nCr form)
1
1
1
1
1
2
3
1
1
1
1
3
1
1
1
1
1
2
3
4
1
3
6
1
4
1
5 10 10 5 1
 Example 2: Express the 8th row of Pascal’s triangle using combinations. Then, evaluate and state as numbers.
 Example 3: Use your result from question 2, along with patters of exponents, to fully expand (𝑥 + 𝑦)7
 Example 4: For each of the following binomial in the form (𝑥 + 𝑦)𝑛 , identify 𝒙 (first term), 𝒚 (second term), and the
number of terms in the expansion:
(a) (5𝑥 + 1)13
1
𝑥
(b) (4𝑥 2 − )7
(c) (3𝑦 − 11𝑥)5
21
Worked Example 1: Fully Expand (2𝑥 − 3𝑦)4 using Pascal’s Triangle and patterns of binomial expansion
Step
: There will be 𝑛 + 1 = 𝟓 terms in the expansion. (𝑛 is the exponent of the given binomial, here 𝑛 = 4) So
draw 5 sets of triple brackets,
Step
1
: Use the 5th row of Pascal’s triangle.
1
This provides the coefficients
1
1
1
1
2
3
4
1
3
6
1
4
1
Step
: Place the appropriate number / term in each bracket
Step
: Powers of 𝑥 go from 𝑛 to 0, while powers of 𝑦 go from 0 to 𝑛. Write each exponent.
Pascal’s
“𝑥", each
coefficient first term
“𝑦", each
second term
(1)(2𝑥)4 (−3𝑥)0 + (4)(2𝑥)3 (−3𝑥)1 + (6)(2𝑥)2 (−3𝑥)2 + (4)(2𝑥)1 (−3𝑥)3 + (1)(2𝑥)0 (−3𝑥)4
Step
: Simplify!
(1)(16𝑥 4 )(1) + (4)(8𝑥 3 )(−3𝑥) + (6)(4𝑥 2 )(9𝑥 2 ) + (4)(2𝑥)(−27𝑥 3 ) + (1)(1)(81𝑥 4 )
𝟏𝟔𝒙𝟒 − 𝟗𝟔𝒙𝟑 𝒚 + 𝟐𝟏𝟔𝒙𝟐 𝒚𝟐 − 𝟐𝟏𝟔𝒙𝒚𝟑 + 𝟖𝟏𝒙𝟒
You Try:
 Example 5: Fully expand (3𝑥 − 𝑦)5
(Answer is on the top of the next page)
22
Answer from previous page: #5.
Worked Example 2: Find the 4th term in the expansion of (𝟐𝒙 − 𝒚)𝟕.
Step
(using the binomial theorem)
: We are going to use the formula:
Identify “𝒌”
We want the 4th term, 𝑡4 – which, we can write in the form 𝑡𝑘+1 as 𝒕𝟑+𝟏 . 𝒌 = 𝟑.
Step
: Identify the other parameters for
𝒌 = 𝟑,
Represents term
we want, always
“one less”
Step
𝒙 = 𝟐𝒙,
First term in
binomial
𝒚 = −𝒚,
𝒏=𝟕
Exponent of binomial
Second term in
binomial
: Substitute all values and simplify!
𝒕𝟑+𝟏 = ⬚𝟕𝑪𝟑 (𝟐𝒙)𝟕−𝟑 (−𝒚)𝟑
𝒕𝟒 = 𝟑𝟓(𝟏𝟔𝒙𝟒 )(−𝒚𝟑 )
= −𝟓𝟔𝟎𝒙𝟒 𝒚𝟑
You Try:
 Example 6: Find the 8th term in the expansion of (1 − 3𝑚)10
 Example 7: Find the 3rd term in the expansion of (4𝑥 2 − 𝑦 3 )5
Answers from #6.
previous page: #7.
Worked Example 3: Find the term involving 𝑥 6 in the expansion of (𝟑𝒙 + 𝟒𝒚)𝟖.
Step
23
(using the binomial theorem)
: We are going to use the formula:
Identify “𝒌”
Notice that in the formula the exponent of 𝒙 is “𝒏 − 𝒌”. So, we want 𝒏 − 𝒌 = 𝟔
𝟖−𝒌=𝟔
𝒌=𝟐
Step
Aside: Knowing that 𝒌 = 𝟐 means…
: Identify the other parameters for
 On our term the exp of 𝑦 will be 2
𝒌 = 𝟐,
Step
𝒙 = 𝟑𝒙,
𝒚 = 𝟒𝒚,
𝒏=𝟖
rd
 Our wanted term is the 3 term in
the expansion of (3𝑥 + 4𝑦)8 .
(𝑘 is always 1 less than the term / 𝑡𝑘+1 )
: Substitute all values and simplify!
𝒕𝟐+𝟏 = ⬚𝟖𝑪𝟐 (𝟑𝒙)𝟖−𝟐 (𝟒𝒚)𝟐
𝒕𝟑 = 𝟐𝟖(𝟕𝟐𝟗𝒙𝟔 )(𝟏𝟔𝒚𝟐 )
= 𝟑𝟔𝟐𝟓𝟗𝟐𝒙𝟔 𝒚𝟐
By the way! Here is (𝟑𝒙 + 𝟒𝒚)𝟖 fully expanded. Notice the 3rd term / the term that involves 𝒙𝟔
I know you – you’re the term
involving 𝑥 6 we just found without
having to do this “fully expanding”!
 Example 8: Find the term involving 𝑥 3 in the expansion of (5𝑥 − 2𝑦)10 .
 Example 9: Find the term involving 𝑦 8 in the expansion of (4𝑥 − 3𝑦 2 )6 .
Answers from #8.
previous page: #9.
 Example 10:
Diploma
Example
 Example 11 In the expansion of (2𝑥 − 5𝑦 2 )6 , what is the coefficient of the term involving 𝑥 3 𝑦 6?
 Example 12
 Example 13
Diploma
Example
Answers on next page #14. 11 520
24
Answers from #10. a=12
previous page: #12. 𝒂 = 𝟒𝟐
𝟏
Worked Example 4: Find the constant term in the expansion of (𝟑𝒙 − 𝒙𝟐 )𝟗 .
Step
#11. -20 000
#13. 124
(using the binomial theorem)
: We are going to use the formula:
Identify “𝒌”
In a constant term there is “no 𝒙”, that is the exponent of 𝑥 is 0.
This time, both terms have an 𝑥 in them. So we start plugging values into the formula to see if
we can determine k:
(3𝑥)9−𝑘 (−1𝑥 −2 )𝑘
𝒕𝒌+𝟏 = ⬚𝟗𝑪𝒌 (𝟑𝒙)𝟗−𝒌 (−
𝟏 𝒌
)
𝒙𝟐
= 39−𝑘 ∗ 𝒙𝟗−𝒌 ∗ (−1)𝑘 ∗ 𝒙−𝟐𝒌
Goal: We want the exponent of 𝒙 to be 0!
Focus on just the 𝑥 terms…
So…simplify the terms involving 𝑥
= 𝒙𝟗−𝒌 ∗ 𝒙−𝟐𝒌
= 𝒙𝟗−𝒌+(−𝟐𝒌)
REMEMBER! We want the
exp of 𝑥 to be 0!
𝟗 − 𝟑𝒌 = 𝟎
𝒌=𝟑
Step
: Phew! Not go back to the formula:
Step
: Substitute all values and simplify!
𝒌 = 𝟑,
Hopefully we’ll get a nice constant
value! (ie, just a number)
𝒕𝟑+𝟏 = ⬚𝟗𝑪𝟑 (𝟑𝒙)𝟗−𝟑 (−
𝒕𝟑 = 𝟖𝟒(𝟕𝟐𝟗𝒙𝟔 )(−
=
−𝟐𝟔𝟐𝟒𝟒𝒙𝟔
𝒙𝟔
= −𝟔𝟏 𝟐𝟑𝟔
 Example 14. Find the constant term in the expansion of
𝟏
𝒙𝟐
)𝟑
𝟏
)
𝒙𝟔
𝒙 = 𝟑𝒙,
𝟏
𝒚 = − 𝒙𝟐 ,
𝒏=𝟗
25