CMA Foundation · Fundamentals of Business Mathematics and Statistics

Permutation and Combinations: formula sheet

Full chapter guide

Key formulas

Multiplication rule: If task 1 can be done in m ways and, for each of these, task 2 can be done in n ways, total ways = m × n; Extends to any number of stages: m × n × p × ... Use when all stages must be completed in sequence.
Addition rule: If task A can be done in m ways or task B in n ways, and both cannot be done together, total ways = m + n; Applies only when the alternatives are mutually exclusive, so no way is counted twice.
Choices with restriction at a stage: Count the restricted stage first, then multiply the remaining stages; For example, a number that must be even: fix the units digit first.
Definition of n!: n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1; Valid for natural numbers n. Example: 4! = 24.
Recursive property: n! = n × (n − 1)!; Use it to expand a factorial just enough to cancel. Also valid as n! = n(n − 1)(n − 2)!, for n ≥ 2.
Zero factorial: 0! = 1; It is a definition. Also 1! = 1.
Values to remember: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040; Knowing these saves time in MCQs.
Ratio of factorials: n! ÷ r! = n × (n − 1) × ... × (r + 1), for n > r; Cancel the smaller factorial; the product has n − r terms.
Not distributive: (m + n)! ≠ m! + n! and (m × n)! ≠ m! × n!; Always compute the bracket first, then take the factorial.
Permutation of n distinct objects taken r at a time: nPr = n! ÷ (n − r)!; Valid for 0 ≤ r ≤ n, with n and r whole numbers. No repetition.
Product form: nPr = n × (n − 1) × (n − 2) × ... (r factors); Fastest for calculation. Write exactly r numbers going down from n.
All objects arranged: nPn = n!; Since 0! = 1, this follows from the main formula.
Special values: nP0 = 1, nP1 = n; One way to choose nothing; n ways to fill one place.
Recurrence link: nPr = n × (n−1)P(r−1); Useful for solving equations such as nPr = k × (n−1)P(r−1).
Arrangements with identical objects: n! ÷ (p! × q! × r! × …); n objects in total, of which p are of one kind, q of another, r of another. Objects that occur once need no division.
Repetition allowed: n^r; Number of ways to fill r positions when each position can be filled in n ways and reuse is allowed.
Circular permutation of distinct objects: (n − 1)!; Applies when only the relative positions matter and clockwise and anticlockwise are counted as different.
Circular permutation when clockwise and anticlockwise are the same: (n − 1)! ÷ 2; Used for necklaces, garlands and similar cases that can be flipped over. Valid for n ≥ 3.
Items always together (linear): (n − k + 1)! × k!; k specified distinct items form one block among n distinct objects.
Items never together: Total arrangements − Arrangements with the items together; Works for both linear and circular cases. Use the matching total and together counts.
Gap method for never together (linear): (n − k)! × (n − k + 1)P k; Arrange the other n − k objects first, then place the k items in separate gaps. There are n − k + 1 gaps.
Combination formula: nCr = n! ÷ [r! × (n − r)!]; Valid for whole numbers with 0 ≤ r ≤ n.
Link with permutation: nCr = nPr ÷ r!, so nPr = nCr × r!; Use when a question gives one and asks for the other.
Complementary property: nCr = nC(n − r); Choosing r to take equals choosing n − r to leave.
Special values: nC0 = 1, nCn = 1, nC1 = n; Exactly one way to choose none or all.
If nCx = nCy: Either x = y or x + y = n; Use to solve equations in r or n.
Pascal's rule: nCr + nC(r − 1) = (n + 1)Cr; Handy for adding two consecutive combinations.
Ratio of consecutive terms: nCr = [(n − r + 1) ÷ r] × nC(r − 1); Useful for ratio-based equations.
Fast expansion: nCr = [n × (n − 1) × … (r factors)] ÷ r!; Write r factors from n downwards on top, r! below.
Combination formula: nCr = n! ÷ [r! × (n − r)!]; Valid for whole numbers with 0 ≤ r ≤ n. Order does not matter.
Symmetry rule: nCr = nC(n − r); Use it to shorten work. For example, 10C8 = 10C2 = 45.
Special values: nC0 = nCn = 1 and nC1 = n; Choosing none or all can be done in exactly one way.
Selections of any number from n distinct items: nC1 + nC2 + ... + nCn = 2ⁿ − 1; This counts selecting at least one item. Each item is either in or out (2 choices), giving 2ⁿ, and you remove the empty selection.
Particular items included or excluded: Particular person included: (n − 1)C(r − 1). Particular person excluded: (n − 1)Cr; For k particular persons always included: (n − k)C(r − k).
Dividing into labelled groups: n! ÷ (a! × b! × c!), where a + b + c = n; Use when the groups are different in identity, such as given to different people.
Dividing into equal unlabelled groups: n! ÷ [(m!)ᵏ × k!], where n = m × k; k groups of m items each, with groups not named. Divide by k! to remove repeated divisions.

Quick revision

Multiplication principle: if tasks happen one after another, multiply the ways.
Addition principle: if tasks are alternatives and cannot happen together, add the ways.
n! = n × (n − 1) × ... × 2 × 1, and 0! = 1.
nPr = n! ÷ (n − r)!, and nPn = n!.
nCr = n! ÷ [r! (n − r)!], and nPr = nCr × r!.
nCr = nC(n − r), so use the smaller r to calculate faster.
nC0 = nCn = 1 and nC1 = n.
Arrangements of n objects where p are alike of one kind and q alike of another: n! ÷ (p! q!).
Items that must stay together: treat them as one object, then multiply by their internal arrangements.
Never together = total arrangements − arrangements with them together.
Circular arrangement of n distinct objects: (n − 1)!.
If order matters it is a permutation; if it does not, it is a combination.

Common mistakes

Adding when the tasks are in sequence Fix: Ask: do I need both? If yes, multiply. Add only for either-or choices.
Multiplying when the options are alternatives Fix: If you pick from only one group, the groups are alternatives, so add.
Taking 0! = 0 Fix: Remember 0! = 1 by definition. It comes from 1! = 1 × 0!.
Writing (m + n)! = m! + n! Fix: Add inside the bracket first. For example, (2 + 3)! = 5! = 120, but 2! + 3! = 8.
Writing nPr as n! ÷ r! instead of n! ÷ (n − r)! Fix: Remember that permutation has only (n − r)! below. Check with 5P1, which must be 5.
Using more or fewer than r factors in the product Fix: Count the factors: 8P3 has three numbers, 8 × 7 × 6 = 336.
Using n! for a word with repeated letters without dividing. Fix: Count each letter first. Divide n! by the factorial of each repeat count.
Forgetting the internal arrangements of a block in a together question. Fix: Always multiply by k! for the k items inside the block, if they are distinct.
Using permutation when order does not matter Fix: Ask: if I swap two chosen items, is the outcome the same? If yes, use nCr.
Forgetting to divide by r! Fix: Remember nCr = nPr ÷ r!. Always keep r! in the denominator.

Exam tips

Before calculating, underline the words 'and', 'or', 'either' and 'then' in the question. They tell you whether to multiply or add.
For 'even number' or 'number greater than' questions, fix the restricted position first. This avoids losing marks.
Check the repetition condition. Questions that say 'repetition allowed' and 'not allowed' give different answers, and both appear as options.
Since there is no negative marking, attempt every question. If unsure, eliminate options that are not whole numbers or are clearly too small or too large.
Memorise factorials up to 7! so you can check options quickly.
Questions often hide a common factorial. Look for it first before multiplying anything.
If an option equals your answer only after using 0! = 0, you have fallen for a trap. Recheck.
For equations with n, reject negative roots and non-integers. This eliminates options fast.