Start by identifying what makes outcomes different
A counting problem asks how many outcomes satisfy given conditions. Before selecting a formula, decide whether order matters, whether objects may repeat, whether any positions are restricted, and whether apparently different arrangements are being treated as equivalent. The same people can form one committee but several officer teams. For example, choosing Ada and Bayo as members is the same selection as choosing Bayo and Ada; assigning Ada as captain and Bayo as secretary is a different outcome from reversing those roles.
The multiplication principle counts successive choices. If a student can choose one of three shirts and one of four trousers, there are 3 × 4 = 12 outfits, provided every shirt can be paired with every trouser. The addition principle combines alternatives that do not overlap. If two permitted categories overlap, remove the double count or partition them into disjoint cases first. Writing what each factor or case means is more reliable than multiplying every number that appears in a question.
Use factorials for distinct arrangements
For a positive integer n, n! = n(n − 1)...2 × 1. The convention 0! = 1 keeps counting formulas consistent with the single empty arrangement. The number of linear arrangements of n distinct objects is n!: choose one of n objects for the first position, n − 1 for the second, and continue. Four distinct books can be ordered in 4! = 24 ways. Here every position is filled and each book is used exactly once.
A permutation of r objects selected and ordered from n distinct objects is nPᵣ = n!/(n − r)!, for integers 0 ≤ r ≤ n. Eight students can fill captain, secretary and treasurer roles in 8 × 7 × 6 = 336 ways when no student holds two roles. The roles make order matter. If repeated assignment were allowed, each role would have eight choices, giving 8³ instead. Do not use a without-replacement formula in a problem that permits repeated objects or repeated people.
Use combinations when order does not matter
A combination of r objects from n distinct objects is nCᵣ = n!/[r!(n − r)!]. Dividing by r! removes the internal orders of each selected group. Choosing three committee members from eight students gives 8C₃ = 56 committees. Each committee would have been counted 3! = 6 times in 8P₃. The symmetry nCᵣ = nCₙ₋ᵣ follows because choosing the included objects also determines the excluded objects. This can simplify a calculation, but it does not change whether order matters.
Combinations apply to selections without replacement from distinct objects. In an elementary model with equally likely selections, they can count favourable and total selections. From a bag of three red and two blue individually distinct counters, the that an unordered draw of two contains two red is. This matches the sequential calculation (3/5)(2/4). Use a consistent sample space: unordered favourable outcomes should not be divided by an ordered total.
Handle repeated identical objects
If n positions are occupied by identical objects in groups of sizes a, b and so on, with a + b + ... = n, the number of distinct linear arrangements is n!/(a!b!...). The word LEVEL has five letters, with two Ls, two Es and one V. Its distinct arrangements number 5!/(2!2!) = 30. Swapping the two visually identical Ls does not produce a different word, which explains the division. Treat physically labelled copies as distinct only if the question makes those labels relevant.
Compare this with a repeated-choice code. A four-digit code using digits 0 through 9 with repetition allowed has 10⁴ possibilities if leading zero is permitted. A four-digit number cannot start with zero, so it has 9 × 10³ possibilities. If repetition is forbidden, the first count becomes 10 × 9 × 8 × 7 for a code and 9 × 9 × 8 × 7 for a four-digit number. The distinction between a code and a number changes the first-position rule.
Count circular arrangements with stated symmetry
For n distinct people around an unlabelled circular table, rotations describe the same seating, so the number of circular arrangements is (n − 1)!. One method fixes a chosen person in one reference position and arranges the remaining n − 1 people around them. Five people therefore have 4! = 24 circular seatings. Clockwise and anticlockwise orders are normally different seatings. Do not divide by another two unless the problem explicitly identifies reflected arrangements as equivalent, as can happen for some necklace models.
If seats are numbered or one seat has a specified physical label, rotations need not be equivalent and the count may instead be n!. Likewise, repeated objects around a circle require more care than blindly dividing a repeated-object linear formula by n, because some arrangements have extra rotational symmetry. Elementary problems here use distinct people and an unlabelled table. State that assumption in the answer rather than applying the circular formula to every object arranged in a ring.
Use blocks and complementary counting for restrictions
For six distinct people in a row with Ada and Bayo together, treat the pair as one block. There are now five objects to arrange in 5! ways, and the pair can be AB or BA in two ways. The total is 2 × 5! = 240. To keep them apart, subtract these from all 6! arrangements: 720 − 240 = 480. The block has internal arrangements as well as an external position, so omitting the factor 2 loses valid outcomes.
For a three-person committee from three girls and four boys with at least two girls, separate the permitted cases. Exactly two girls and one boy gives 3C₂ × 4C₁ = 12; three girls gives 3C₃ = 1. The total is 13. “At least” and “exactly” different case lists. Self-check: find 5!, choose two from six, arrange three officers from six, and arrange five distinct people around an unlabelled circle. Answers are 120, 15, 120 and 24. Check whether every outcome is counted once and whether any restriction affects an early choice or a later one. These questions establish the model before arithmetic begins.