Look for this. When a finished object needs one choice at each of several stages, draw one slot per stage. If an early choice changes later options, split into disjoint cases.
Multiply the number of choices along each branch, since each starting choice pairs with every allowed next choice. Add counts only across branches that cannot overlap.
Worked example
A rental shop offers 4 bike frames, 3 helmets and 2 light kits. One frame is too small for the larger light kit. How many complete frame–helmet–light packages work?
The three unrestricted frames give 3 × 3 × 2 = 18 packages. The small frame gives 1 × 3 × 1 = 3. The cases have different frames, so add to get 21.
Try it yourself
A locker label has three different digits from 0 to 9. The first cannot be 0 and the last must be even. How many labels can the school issue?
Check your reasoning
If the last digit is 0, there are 9 choices for the first and 8 for the middle, giving 72. If the last digit is 2, 4, 6 or 8, there are 4 choices for it, 8 nonzero choices for the first and 8 for the middle, giving 256. The total is 328 labels.
Outcome PC2
Count arrangements when positions matter
Look for this. Swap two entries. If the swap changes the result, positions matter. If two entries look identical, swapping those copies creates no new arrangement.
For r different positions filled from n distinct items, use nPᵣ = n!/(n − r)!. For a full arrangement with identical entries, divide n! by the factorial of each repeated count.
Worked example
A festival has 7 distinct films and 3 screening times. One documentary can only run in the last time slot if selected. How many schedules have three different films?
Without the documentary, arrange 3 of the other 6 in 120 ways. With the documentary last, arrange 2 of the other 6 in 30 ways. The total is 150 schedules.
Try it yourself
On a grid, a courier goes exactly 4 blocks east and 3 blocks north. The first block must be east. How many different shortest routes are available?
Check your reasoning
Fix the first E. Each route is now an arrangement of E,E,E,N,N,N. Ordering six labelled moves would give 720, but reordering the three E labels and the three N labels repeats each route 3! × 3! times. There are 6!/(3!3!) = 20 routes.
Outcome PC3
Count a team when membership matters
Look for this. A rearrangement of the same members is still the same team. Requirements such as ‘at least one’ suggest counting and removing the forbidden case.
Choosing r of n distinct items gives C(n, r) = n!/[r!(n − r)!]. The r! removes the orders of each chosen group. For separate categories, multiply independent selections.
Worked example
A club has 5 experienced members and 4 new members. How many four-person crews include at least one new member?
There are C(9, 4) = 126 unrestricted crews. An all-experienced crew has C(5, 4) = 5 possibilities. Subtract to get 121 eligible crews.
Try it yourself
Six prints and five sculptures are available for a show. Choose exactly two of each. Two particular prints must appear together or both stay out. Count the selections.
Check your reasoning
If the pair appears, the prints are fixed. If it stays out, choose two of the other four in C(4, 2) = 6 ways. Either group of prints can accompany C(5, 2) = 10 sculpture pairs, so (1 + 6) × 10 = 70 selections.
Outcome PC4
Find a binomial term by counting its choices
Look for this. A power of a two-term expression represents identical factors. Each term in the expansion records how many times you picked the second entry.
Picking the second entry k times in n factors gives C(n, k) ways, so its contribution is C(n, k) a^(n − k)b^k. For a full expansion, let k run from 0 to n. Keep any negative sign inside b.
Worked example
Expand (2x − 1)⁴ and explain why the signs alternate.
The fourth-power coefficients are 1, 4, 6, 4, 1. Attach (2x)^(4 − k)(−1)^k to each. The odd k choices give negative terms, so the result is 16x⁴ − 32x³ + 24x² − 8x + 1.
Try it yourself
A student wants just the coefficient of x⁵ in (x² + 3x)⁴. Determine it without expanding everything and explain how you chose the term.
Check your reasoning
If 3x is chosen k times, the power of x is 2(4 − k) + k = 8 − k. To reach x⁵ we need k = 3. There are C(4, 3) = 4 placements of the three 3x factors. Each contributes 3³ = 27, so the coefficient is 108.
Worked notes and diagrams
1
The words, first
The idea: counting problems are wordy on purpose, and the answer usually hides in one of these terms.
Order and repetition pick the tool. Restrictions then decide which box you fill first.
Word
What it means
Outcome
One complete result — one code, one arrangement, one committee.
Fundamental counting principle
If one stage happens m ways and the next n ways, together they happen m × n ways.
Permutation
A selection where order matters, so ABC and CBA are different outcomes.
Combination
A selection where order does not matter, so ABC and CBA are the same outcome.
Factorial
n! = n × (n − 1) × … × 1, the arrangements of n different items. 0! = 1.
Restriction
A condition limiting an outcome: two people together, a code starting with a vowel.
Case
A branch counted separately, then added to the others.
Complement
The outcomes you do not want — count everything and subtract them.
Repeated items
Identical items in an arrangement. Divide by the factorial of each repeat.
Binomial expansion
The expanded form of (x + y)ⁿ, whose coefficients are ₙCₖ.
General term
Tₖ₊₁ = C(n, k) xⁿ⁻ᵏ yᵏ. The term number is k + 1. Each exponent counts how many factors supply its variable.
2
The fundamental counting principle
The idea: if one decision can be made in m ways and a second in n ways, the two together can be made in m × n ways. Draw a box for each decision, write the number of choices in it, and multiply.
This is the foundation everything else sits on. The trick is not the multiplying — it is noticing how many separate decisions there actually are.
Worked example
A restaurant special includes one appetizer, one main and one dessert. There are 2 appetizers, 5 mains and 3 desserts. How many different meals are possible?
Three decisions, so three boxes:
One box per decision. Multiply across the boxes: 2 × 5 × 3 = 30.
When choices run out
If something cannot be reused, the number in each box drops by one as you go. A 4-digit PIN using only the digits 1–5 with no repeats is 5 × 4 × 3 × 2 = 120, not 5⁴.
Watch out: adding instead of multiplying. With 3 shirts and 4 pairs of pants, 3 + 4 = 7 counts the pieces of clothing. The outfits are 3 × 4 = 12.
3
Permutations
The idea: order matters. ₙPᵣ = n! ÷ (n − r)!. To arrange n objects where some are identical, divide n! by the factorial of each repeat count.
A permutation is an arrangement. Swapping two things gives you a different permutation — which is exactly what makes it different from a combination.
Factorials first
n! means multiply n by every whole number below it, down to 1. So 5! = 5 × 4 × 3 × 2 × 1 = 120. It is not 5 × 5, and it does not stop partway.
Worked example
Evaluate ₈P₂.
The formula gives 8! ÷ 6!. Written out, everything from 6 downwards appears on the top and the bottom, so it cancels:
The matching factors cancel, leaving 8 × 7. Shortcut: ₙPᵣ is just r factors counting down from n.
When letters repeat
BOOK has 4 letters, so 4! = 24 — except the two O's are identical, so every word got counted 2! times. Divide it out: 24 ÷ 2 = 12. MISSISSIPPI has I four times, S four times and P twice, so it is 11! ÷ (4! 4! 2!) = 34 650.
Watch out: treating repeated letters as if they were different. If swapping two of them gives the same word, you have counted it twice.
4
Combinations
The idea: order does not matter. ₙCᵣ = n! ÷ [(n − r)! r!]. Handy link: ₙCᵣ = ₙPᵣ ÷ r!, because each group can be ordered r! ways.
A combination is a group. Picking Ana then Ben gives the same group as picking Ben then Ana, so it should only be counted once.
Try it: does order matter?
Every arrangement is listed, grouped by which people it uses. Switch "order doesn't matter" on and watch each row of arrangements collapse into a single group — that collapse is the division by r!.
Worked example
Evaluate ₅C₂ — choosing 2 things from 5.
₅C₂ = 5! ÷ (3! × 2!) = 120 ÷ 12 = 10. You can see all ten at once by drawing 5 points and joining every pair:
Choosing 2 of 5 is like drawing a line between 2 of 5 points. There are 10 lines: AB AC AD AE BC BD BE CD CE DE.
Choosing from two groups
If a team of 4 needs exactly 2 girls from 7 and 2 boys from 5, choose each group separately and multiply: ₇C₂ × ₅C₂ = 21 × 10 = 210.
Watch out: using a permutation when order does not matter. ₅P₂ = 20 but ₅C₂ = 10 — in general ₙPᵣ = r! × ₙCᵣ. Thus ₙCᵣ ≤ ₙPᵣ, with equality at r = 0 or r = 1.
5
The binomial theorem
The idea: (a + b)ⁿ has n + 1 terms. The coefficients are row n of Pascal's triangle, which are also ₙC₀, ₙC₁, …, ₙCₙ. The general term is ₙCₖ aⁿ⁻ᵏ bᵏ.
Expanding a bracket to a power by hand gets miserable fast. Pascal's triangle hands you the coefficients for free.
Each number is the sum of the two above it. Rows start at row 0, so row 3 gives (a + b)³ = a³ + 3a²b + 3ab² + b³.
Finding one particular term
You rarely need the whole expansion — usually just one term. Use ₙCₖ aⁿ⁻ᵏ bᵏ and solve for k.
For the coefficient of x³ in (x + 2)⁵: you need x³, so n − k = 3, meaning k = 2. The term is ₅C₂ x³ 2² = 10 × 4 = 40x³.
For the constant term in (x + 1/x)⁶: the powers give x⁶⁻ᵏ × x⁻ᵏ = x⁶⁻²ᵏ. A constant needs the power to be zero, so k = 3, and the term is ₆C₃ = 20.
Watch out: forgetting to raise the whole term to the power. In (2x + 3)³ the first term is (2x)³ = 8x³, not 2x³. Bracket it before you cube it.
Which one do I need?
Ask one question: if I swap two of these around, is it different?
Yes, it's different → order matters → permutation. Look for roles, rankings, positions, passwords, or a row.
No, it's the same → order does not matter → combination. Look for teams, committees, groups, hands of cards, or toppings.
Several separate decisions in a row → multiply the boxes → counting principle.
A bracket raised to a power → binomial theorem.
Gold, silver and bronze is a permutation. Three students going on a field trip is a combination. Same numbers, different answer.
Practice by topic
Work on paper, choose an answer, then compare your method with the explanation.
Fundamental counting principle · 6 questions
Practice 1
Maya has 3 shirts and 4 pairs of pants. How many different outfits (1 shirt and 1 pair of pants) can she make?
24
34
7
12
Answer and explanation
12
One choice AND another means multiply: 3 × 4 = 12. Adding to get 7 counts the pieces of clothing, not the outfits.
Practice 2
A restaurant special includes 1 appetizer, 1 main course and 1 dessert. There are 2 appetizers, 5 main courses and 3 desserts. How many different meals are possible?
15
30
10
60
Answer and explanation
30
One box per decision, then multiply across: 2 × 5 × 3 = 30. Getting 10 or 15 means one of the three courses was left out.
Practice 3
A coin is flipped 4 times. How many different sequences of heads and tails are possible?
16
4
8
24
Answer and explanation
16
Each flip has 2 outcomes and there are 4 flips, so 2⁴ = 16. 8 is 2³, the answer for only 3 flips — use one factor of 2 per flip.
Practice 4
A lock code uses 3 digits (0–9), and digits may be repeated. How many codes are possible?
30
720
999
1000
Answer and explanation
1000
Repeats are allowed, so every position keeps all 10 digits: 10 × 10 × 10 = 1000 (the codes 000 to 999). 999 forgets 000.
Practice 5
A 4-digit PIN is made using only the digits 1, 2, 3, 4 and 5, with no digit repeated. How many PINs are possible?
120
625
24
20
Answer and explanation
120
No repeats, so the choices drop by one each time: 5 × 4 × 3 × 2 = 120. 625 is 5⁴, which would allow repeats.
Practice 6
How many 4-digit numbers can be formed from the digits 1–9 if no digit is repeated?
504
36
6561
3024
Answer and explanation
3024
Order matters and repeats are banned: ₉P₄ = 9 × 8 × 7 × 6 = 3024. 6561 is 9⁴, which would allow repeats.
Permutations · 10 questions
Practice 1
What is the value of 5! ?
60
25
720
120
Answer and explanation
120
5! = 5 × 4 × 3 × 2 × 1 = 120. It is not 5 × 5, and it does not stop partway — keep multiplying all the way down to 1.
Practice 2
In how many ways can 4 different books be arranged in a row on a shelf?
4
16
24
256
Answer and explanation
24
All 4 books are used and order matters, so it is 4! = 4 × 3 × 2 × 1 = 24. 256 is 4⁴, which would let one book sit in more than one spot.
Practice 3
Evaluate ₈P₂.
28
56
64
16
Answer and explanation
56
ₙPᵣ is r factors counting down from n, so ₈P₂ = 8 × 7 = 56. 28 is ₈C₂, which ignores order.
Practice 4
A club with 6 members must choose a president and a vice-president. How many ways can this be done?
30
15
12
36
Answer and explanation
30
The two positions are different, so order matters: 6 × 5 = 30. 15 treats the pair as a group with no roles, which would be a combination.
Practice 5
How many different arrangements can be made using all the letters of the word BOOK?
6
12
4
24
Answer and explanation
12
4 letters give 4! = 24, but the two O's are identical, so every word was counted 2! times: 24 ÷ 2 = 12.
Practice 6
How many different arrangements can be made using all the letters of the word BANANA?
120
60
720
360
Answer and explanation
60
6 letters give 6! = 720, but A repeats 3 times and N twice: 720 ÷ (3! × 2!) = 720 ÷ 12 = 60.
Practice 7
In how many ways can 5 people sit in a row if two particular people must sit together?
24
96
120
48
Answer and explanation
48
Glue the pair into one block, leaving 4 items to arrange: 4! = 24. The pair can sit in 2 orders inside the block, so 24 × 2 = 48.
Practice 8
Solve for n: ₙP₂ = 42
8
6
21
7
Answer and explanation
7
ₙP₂ = n(n − 1), so n(n − 1) = 42. Two consecutive numbers multiplying to 42 are 7 and 6, so n = 7.
Practice 9
How many different arrangements can be made using all the letters of the word MISSISSIPPI?
138 600
34 650
39 916 800
69 300
Answer and explanation
34 650
11 letters with I ×4, S ×4, P ×2: 11! ÷ (4! × 4! × 2!) = 39 916 800 ÷ 1152 = 34 650. Every repeated letter needs its own factorial on the bottom.
Practice 10
Using the letters A, B, C, D, E exactly once each, how many arrangements have no two vowels next to each other?
48
120
72
36
Answer and explanation
72
Place the 3 consonants first: 3! = 6. That makes 4 gaps; choose 2 for the vowels and order them: ₄C₂ × 2! = 12. Then 6 × 12 = 72.
Combinations · 10 questions
Practice 1
Evaluate ₅C₂.
10
7
25
20
Answer and explanation
10
₅C₂ = 5!/(3! × 2!) = 10. Ordering the two choices gives ₅P₂ = 2! × 10 = 20. In general ₙCᵣ ≤ ₙPᵣ, with equality when r is 0 or 1.
Practice 2
Which situation should be solved using a combination instead of a permutation?
Arranging 3 photos in a row on a wall
Creating a 3-letter password
Choosing 3 students from a class of 25 to attend a field trip
Awarding gold, silver and bronze medals to 3 of 10 runners
Answer and explanation
Choosing 3 students from a class of 25 to attend a field trip
The same 3 students go on the trip no matter who is picked first, so order does not matter. Medals, passwords and rows all change meaning when you reorder them.
Practice 3
A pizza shop offers 6 toppings. How many different pizzas with exactly 3 different toppings can be made?
216
20
120
18
Answer and explanation
20
The order toppings are added does not change the pizza: ₆C₃ = 720 ÷ (6 × 6) = 20. 120 is ₆P₃, which counts the same pizza 6 times.
Practice 4
A committee of 2 people is chosen from a group of 10. How many different committees are possible?
20
45
90
100
Answer and explanation
45
No roles on the committee, so order does not matter: ₁₀C₂ = (10 × 9) ÷ 2 = 45. 90 is ₁₀P₂, counting "Ana and Ben" separately from "Ben and Ana".
Practice 5
At a meeting, each of 6 people shakes hands once with every other person. How many handshakes occur?
12
30
15
36
Answer and explanation
15
A handshake needs 2 people and A-with-B is the same as B-with-A: ₆C₂ = 30 ÷ 2 = 15. 30 counts each handshake twice.
Practice 6
How many different 5-card hands can be dealt from a standard 52-card deck?
2 598 960
270 725
311 875 200
22 100
Answer and explanation
2 598 960
A hand is a group, not an order: ₅₂C₅ = 2 598 960. The larger number 311 875 200 is ₅₂P₅, which counts the same hand in every dealing order.
Practice 7
A team of 4 is chosen from 7 girls and 5 boys. How many teams have exactly 2 girls and 2 boys?
70
420
210
495
Answer and explanation
210
Choose each group separately, then multiply: ₇C₂ × ₅C₂ = 21 × 10 = 210. Adding instead of multiplying is the usual slip here.
Practice 8
A committee of 3 is chosen from 8 people, and one of the three is then named chair. How many outcomes are possible?
336
168
512
56
Answer and explanation
168
Choose the group, then pick the chair: ₈C₃ × 3 = 56 × 3 = 168. You can also pick the chair first: 8 × ₇C₂ = 8 × 21 = 168.
Practice 9
A committee of 5 is chosen from 6 men and 4 women. How many committees contain at least 3 women?
66
120
60
126
Answer and explanation
66
"At least 3" splits into exactly 3 and exactly 4: ₄C₃ × ₆C₂ + ₄C₄ × ₆C₁ = 4 × 15 + 1 × 6 = 66. Each case is counted separately, then added.
Practice 10
Solve for n: ₙC₂ = 28
8
28
7
14
Answer and explanation
8
ₙC₂ = n(n − 1) ÷ 2, so n(n − 1) = 56. Consecutive numbers multiplying to 56 are 8 and 7, so n = 8.
Binomial theorem · 10 questions
Practice 1
How many terms are in the expansion of (x + y)⁶?
5
12
7
6
Answer and explanation
7
An expansion of (a + b)ⁿ has n + 1 terms, so 6 + 1 = 7. Counting k from 0 to 6 gives 7 values, not 6.
Practice 2
Which row of Pascal's triangle gives the coefficients in the expansion of (a + b)³?
1 3 1
1 2 1
1 3 3 1
1 4 6 4 1
Answer and explanation
1 3 3 1
Row 3 is 1 3 3 1, matching (a + b)³ = a³ + 3a²b + 3ab² + b³. Rows start at row 0, so 1 4 6 4 1 is row 4.
Practice 3
What is the expansion of (x + 1)²?
x² + x + 1
x² + 2x + 1
x² + 1
2x + 2
Answer and explanation
x² + 2x + 1
(x + 1)(x + 1) = x² + x + x + 1 = x² + 2x + 1. Squaring a bracket is not the same as squaring each term.
Practice 4
In the expansion of (a + b)⁴, what is the coefficient of the middle term, a²b²?
2
6
8
4
Answer and explanation
6
Row 4 of Pascal's triangle is 1 4 6 4 1, and the middle number is 6. It also equals ₄C₂, choosing which 2 of the 4 brackets give a b.
Practice 5
What is the first term in the expansion of (2x + 3)³?
8x³
2x³
27
6x³
Answer and explanation
8x³
The first term takes 2x from every bracket: (2x)³ = 2³x³ = 8x³. Writing 2x³ cubes only the x — bracket the whole 2x first.
Practice 6
What is the coefficient of x³ in the expansion of (x + 2)⁵?
10
20
80
40
Answer and explanation
40
The general term is ₅Cₖ x⁵⁻ᵏ 2ᵏ. For x³ take k = 2: ₅C₂ × 2² = 10 × 4 = 40. Forgetting to raise the 2 leaves just 10.
Practice 7
How many terms are in the expansion of (2x − 3y)⁸?
16
8
9
7
Answer and explanation
9
The number of terms is n + 1 = 9. The coefficients 2 and −3 change the numbers in front but never the number of terms.
Practice 8
What is the middle term in the expansion of (x + 2)⁶?
20x³
240x²
160x³
64x³
Answer and explanation
160x³
7 terms means the 4th is the middle, with k = 3: ₆C₃ x³ 2³ = 20 × 8 × x³ = 160x³.
Practice 9
What is the constant term in the expansion of (x + 1/x)⁶?
15
6
20
64
Answer and explanation
20
The general term gives x⁶⁻ᵏ × x⁻ᵏ = x⁶⁻²ᵏ. A constant needs 6 − 2k = 0, so k = 3 and the term is ₆C₃ = 20.
Practice 10
In the expansion of (2x − y)⁵, what is the coefficient of x²y³?
−80
−40
80
40
Answer and explanation
−40
Take k = 3: ₅C₃ (2x)² (−y)³ = 10 × 4x² × (−y³) = −40x²y³. The minus sign survives because it is raised to an odd power.
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