This topic begins with a deceptively simple question: how many ways? How many ways to arrange five books on a shelf, to pick a team of three from a class, to choose a four-digit code. Counting these possibilities is the foundation of the whole subject — and, as the next page shows, it is also the engine behind probability, since the chance of something is built from counting the ways it can happen.
The work splits by one question that you must ask every time: does order matter? Arranging books in a line (order matters) is a different count from choosing a team (order doesn't). These two cases have two different tools — permutations and combinations — and telling them apart is the single most important skill on this page. We will build both from one starting principle.
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The multiplication principle
Here is the idea everything rests on. If you make a sequence of choices, and the first can be done in m ways and the second in n ways, then the two together can be done in m × n ways. Each of the m first choices branches into n second choices, so the totals multiply.
Picture choosing an outfit from 4 shirts and 3 pairs of trousers. Each shirt can pair with any of the 3 trousers, giving 3 outfits per shirt; with 4 shirts that is 4 × 3 = 12 outfits in all. A tree makes the branching visible.
Two shirts, each branching to three trousers: 2 × 3 = 6 outfits. With 4 shirts it would be 4 × 3 = 12.
Multiplication principle. If a task is a sequence of independent choices with m, n, p, … options at each stage, the total number of ways is the product m × n × p × … Multiply the options at each step.
Worked example · a sequence of choices
A restaurant offers 3 starters, 4 main courses, and 2 desserts. How many different three-course meals are possible?
There are three stages, with 3, 4, and 2 options. By the multiplication principle, multiply them.
3 × 4 × 2 = 24.
So 24 different meals are possible. Each starter opens onto 4 mains, and each of those onto 2 desserts, so the choices multiply rather than add.
Try it! · Check-in question
A password is one letter (26 options) followed by one digit (10 options). How many such passwords are possible?
There are two stages: 26 choices for the letter, then 10 for the digit. By the multiplication principle, 26 × 10 = 260. The answer is C. Option A adds the options (26 + 10 = 36) instead of multiplying them. Option B uses only the digits 1 to 9 (26 × 9 = 234), but 0 is also a digit. Option D doubles the count (2 × 260 = 520), as if the digit could also come first. Here the order is fixed: letter first, then digit.
Try it! · Check-in question
Each student joins one sports club and one art club. There are 6 sports clubs, and there are 24 possible sports-and-art pairs. How many art clubs are there?
By the multiplication principle, (sports clubs) × (art clubs) = (number of pairs). So 6 × (art clubs) = 24, and there are 24 ÷ 6 = 4 art clubs. The answer is A. Option B subtracts (24 − 6 = 18), as if the choices added. Option C adds (24 + 6 = 30). Option D multiplies (24 × 6 = 144) when it should divide. Check: 6 × 4 = 24 ✓.
Factorials
A special case of the multiplication principle gives us a handy piece of notation. Suppose you arrange n distinct objects in a row: there are n choices for the first position, then n − 1 left for the second, n − 2 for the third, and so on down to 1. The product n × (n−1) × … × 2 × 1 comes up so often it gets its own symbol: n!, read "n factorial."
Factorial. n! = n × (n−1) × (n−2) × … × 2 × 1. It counts the number of ways to arrange n distinct objects in order. By convention 0! = 1. For example 4! = 4 × 3 × 2 × 1 = 24.
Worked example · arranging in a row
In how many ways can 4 different books be arranged on a shelf?
There are 4 choices for the leftmost book, then 3 remaining for the next, 2 for the next, and 1 for the last.
By the multiplication principle this is 4 × 3 × 2 × 1 = 4!.
Evaluate: 4! = 24.
So 24 arrangements. Each time you place a book, one fewer remains — which is exactly what the factorial captures.
Try it! · Check-in question
The value of 5! is ( ).
5! = 5 × 4 × 3 × 2 × 1 = 120. The answer is D. Option A squares the 5 (5 × 5 = 25). Option B stops after two factors (5 × 4 = 20). Option C adds instead of multiplying (5 + 4 + 3 + 2 + 1 = 15). The factorial multiplies all the way down to 1, giving 120.
Permutations: when order matters
Often you don't arrange all the objects — you choose and arrange just some of them. Picking a 1st, 2nd, and 3rd place from 6 runners, or filling 2 ranked positions from 5 candidates, is a permutation: an ordered selection. The multiplication principle still drives it — you just stop early.
For 1st, 2nd, 3rd from 6 runners: 6 choices for gold, then 5 for silver, then 4 for bronze, giving 6 × 5 × 4 = 120. Notice this is the top of a factorial that stops after 3 factors. Writing it with factorials, it equals 6! / 3! — the full 6! with the unwanted tail 3! divided away.
Permutation (order matters). The number of ways to choose and arrange r objects from n distinct objects is P(n, r) = n! / (n − r)!. It counts ordered selections — where rearranging the same objects gives a different outcome.
Worked example · an ordered selection
From 5 candidates, a president and a vice-president are chosen (two distinct ranked roles). How many outcomes are possible?
Order matters here: "Ana president, Ben vice" differs from "Ben president, Ana vice." So this is a permutation of 2 from 5.
By the multiplication principle: 5 choices for president, then 4 for vice-president, giving 5 × 4 = 20.
So 20 outcomes. The two ranked roles make order count, which is the hallmark of a permutation.
Try it! · Check-in question
A PIN code has 4 digits (0–9), and all 4 digits must be different. How many PIN codes are possible?
Order matters in a PIN: 1234 and 4321 are different codes. There are 10 choices for the first digit, then 9 for the second (it must be different), 8 for the third and 7 for the fourth: 10 × 9 × 8 × 7 = 5040. This is P(10, 4). The answer is B. Option A, 10⁴ = 10000, allows repeated digits. Option C, 9 × 9 × 8 × 7 = 4536, assumes the first digit cannot be 0, but a PIN can start with 0. Option D multiplies 10 × 4 = 40 (number of digits × length of the code).
Try it! · Check-in question
Five people sit in a row of 5 seats. Ann and Ben must sit in the two end seats. How many seatings are possible?
Deal with the condition first. Ann and Ben take the two end seats in 2 ways (Ann left and Ben right, or the other way round). The other 3 people fill the 3 middle seats in 3! = 6 ways. Total: 2 × 6 = 12. The answer is D. Option A, 5! = 120, ignores the condition. Option B, 3! = 6, forgets that Ann and Ben can swap ends. Option C, 4! = 24, puts Ann at one end and then arranges the other 4 people freely, so Ben is not kept at the other end.
Combinations: when order doesn't matter
Now the strand the CSCA tests most. Often the order of a selection is irrelevant: a committee of 3 is the same committee however you list its members; a hand of cards is the same hand in any order. Choosing without regard to order is a combination.
The link to permutations is exact and worth seeing. A permutation counts ordered selections; a combination counts the same selections but treats all orderings of one group as identical. Since r chosen objects can themselves be arranged in r! orders, every single combination was counted r! times among the permutations. So to get combinations, divide the permutation count by r!.
From permutations to combinations. C(n, r) = P(n, r) / r! = n! / (r!(n − r)!). The division by r! removes the duplicate orderings: each unordered group of r was counted r! times as ordered arrangements, so dividing collapses them to one.
Combination (order doesn't matter). The number of ways to choose r objects from n distinct objects, ignoring order, is C(n, r) = n! / (r!(n − r)!), often written ⁿCᵣ or "n choose r." Rearranging a chosen group gives the same outcome.
Worked example · choosing a committee
How many ways are there to choose a committee of 3 people from 8?
A committee is unordered — the same three people form the same committee in any order — so this is a combination, C(8, 3).
Apply the formula: C(8, 3) = 8! / (3! × 5!).
Compute the top three factors over 3!: (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6.
Divide: 336 / 6 = 56.
So 56 committees. Had the three roles been distinct (chair, secretary, treasurer), order would matter and we would use P(8, 3) = 336 — exactly 3! = 6 times larger.
The one question that decides it: does order matter? If rearranging the chosen objects makes a different outcome (ranked medals, distinct roles, seats in a row), it is a permutation — divide by nothing extra. If rearranging makes the same outcome (a committee, a team, a hand, a subset), it is a combination — divide by r!. Choosing "a committee" but computing a permutation (the most common error) overcounts by exactly r!.
Try it! · Check-in question
A group has 4 boys and 3 girls. In how many ways can you choose 2 boys and 1 girl (order irrelevant)?
Do it in two stages. Choose the boys: C(4, 2) = (4 × 3)/2 = 6 ways. Choose the girl: C(3, 1) = 3 ways. By the multiplication principle, 6 × 3 = 18. The answer is C. Option A, C(7, 3) = 35, chooses any 3 people and ignores the "2 boys and 1 girl" rule. Option B adds the two stages (6 + 3 = 9) instead of multiplying. Option D uses P(4, 2) = 12 for the boys (12 × 3 = 36), but the two boys are not ranked, so order does not matter.
Try it! · Check-in question
In a group of 6 people, everyone shakes hands once with everyone else. How many handshakes occur?
A handshake is unordered (A with B is the same as B with A), so count pairs: C(6, 2) = (6 × 5)/2 = 15. The answer is A. Option C is P(6, 2) = 30, which counts "A shakes B" and "B shakes A" as two different handshakes, so every handshake is counted twice. Option B, 6 × 6 = 36, also lets each person shake their own hand. Option D, 5, is the number of hands one person shakes, not the total. The number of unordered pairs is 15.
Two small facts about combinations are worth knowing. Choosing 0 objects, or all n, can each be done exactly one way: C(n, 0) = 1 and C(n, n) = 1. And there is a pleasing symmetry — choosing which r to include is the same as choosing which n − r to leave out — so C(n, r) = C(n, n − r). This symmetry can save arithmetic: C(10, 8) is easier as C(10, 2).
Try it! · Check-in question
Using the symmetry C(n, r) = C(n, n − r), the value of C(6, 4) equals ( ).
By symmetry C(6, 4) = C(6, 2) = (6 × 5)/2 = 15. The answer is D. Option A multiplies n by r (6 × 4 = 24) instead of using the formula. Option B is P(6, 2) = 30: it forgets to divide by 2!. Option C is P(6, 4) = 6 × 5 × 4 × 3 = 360: the ordered count for choosing 4, without dividing by 4! = 24. Choosing 4 to keep is the same as choosing 2 to leave out, so the answer is 15.
Your turn! · Practice questions
Practice · Question 1
Three-digit numbers are made from the digits 1, 2, 3, 4, 5, and digits may repeat. How many of these numbers are even?
Fill the last digit first, because it carries the condition: an even number ends in 2 or 4, so 2 ways. The first and middle digits can each be any of the 5 digits, because repeats are allowed. Total: 5 × 5 × 2 = 50. The answer is B. Option A, 5³ = 125, ignores "even". Option C, 4 × 3 × 2 = 24, does not allow repeated digits, but the question allows them. Option D, 5 × 2 = 10, leaves out the middle digit.
Practice · Question 2
The value of 8! / 6! is ( ).
Write the larger factorial so that the smaller one appears inside it: 8! = 8 × 7 × 6!. Then 8!/6! = (8 × 7 × 6!)/6! = 8 × 7 = 56. The answer is A. Option B, 8 × 7 × 6 = 336, keeps one factor too many: the 6 is part of 6!, so it cancels. Option C, (8 − 6)! = 2! = 2, subtracts inside the factorial, but factorials do not divide that way. Option D, 8/6 = 4/3, cancels the "!" signs as if they were common factors.
Practice · Question 3
A coach picks 4 of 6 runners for a relay race and also decides who runs the 1st, 2nd, 3rd and 4th part. How many ways can the coach do this?
The running order matters, so this is a permutation: 6 choices for the 1st part, then 5, 4 and 3, giving P(6, 4) = 6 × 5 × 4 × 3 = 360. The answer is C. Option A, C(6, 4) = 15, only chooses the team and forgets the running order. Option B, 4! = 24, only orders 4 runners and forgets that they are chosen from 6. Option D, 6⁴ = 1296, lets the same runner run more than one part.
Practice · Question 4
A team of 3 is chosen from 5 boys and 4 girls. How many teams contain at least one girl?
Every team either has no girl or has at least one girl. So count all teams and remove the teams with no girl. All teams: C(9, 3) = (9 × 8 × 7)/6 = 84. Teams with no girl (3 boys): C(5, 3) = 10. So 84 − 10 = 74. The answer is B. Option A, 84, ignores the condition. Option C, 10, is the number of all-boy teams, which must be removed, not kept. Option D picks "the girl" first (4 ways) and then any 2 of the other 8 people (C(8, 2) = 28), giving 4 × 28 = 112. This counts a team with two girls twice, once for each girl, so it is too big.
Practice · Question 5
An octagon has 8 corners. How many diagonals does it have? (A diagonal joins two corners that are not next to each other.)
Any 2 of the 8 corners give one segment, and order does not matter, so there are C(8, 2) = (8 × 7)/2 = 28 segments. Of these, 8 are sides of the octagon, not diagonals. So there are 28 − 8 = 20 diagonals. The answer is D. Option A, 28, forgets to remove the 8 sides. Option B, 8 × 5 = 40, joins each corner to its 5 non-neighbouring corners, but then counts every diagonal twice (once from each end). Half of 40 is 20. Option C, P(8, 2) = 56, treats segment AB and segment BA as different.
Practice · Question 6
How many subsets with exactly 3 elements does the set {a, b, c, d, e, f, g} have?
A subset has no order: {a, b, c} and {c, a, b} are the same subset. So count combinations: C(7, 3) = (7 × 6 × 5)/(3 × 2 × 1) = 210/6 = 35. The answer is A. Option B, P(7, 3) = 210, counts each subset 3! = 6 times, once for each order. Option C multiplies n by r (7 × 3 = 21). Option D, 2⁷ = 128, counts all the subsets of every size, not only those with 3 elements.
Things to remember
Concept
What to know
Multiplication principle
A sequence of choices with m, n, p, … options gives m × n × p × … total ways. Choices in sequence multiply.
Factorial
n! = n × (n−1) × … × 1, the number of orderings of n objects. 0! = 1.
Permutation (order matters)
P(n, r) = n!/(n−r)! — ordered selections of r from n. Ranked positions, distinct roles, seats in a row.
