Grouping and Distribution of Objects
On this page
Direct answer
Distribution problems ask how many ways objects can land in groups, and the answer changes completely with two distinctions: objects distinct or identical, groups labelled or unlabelled. Distinct objects into labelled groups of sizes n₁, n₂, ..., n_k: n!/(n₁!n₂!...n_k!). Distinct into unlabelled groups of equal size: divide further by the factorial of the number of equal-sized groups. Identical objects into labelled groups allowing empties: C(n + r − 1, r − 1) for n objects into r groups; identical into unlabelled groups is partition counting. The NCERT staple — 12 students split into three equal teams — lands at 12!/(4!·4!·4!·3!) = 5775, and the 3! in the denominator is the entire examination.
What you must remember
- Distinct into labelled groups: sizes n₁, ..., n_k fixed gives n!/(n₁!...n_k!); if sizes are free (any group may take any number), the answer is k^n.
- Equal-size correction: when groups are unlabelled and some sizes coincide, divide by the factorial of the count of coinciding sizes — teams of 4, 4, 4 need the extra ÷3!.
- Distinct into unlabelled groups: split n distinct objects into k non-empty unlabelled groups = Stirling partition counts; for unequal sizes n₁ ≠ n₂ ≠ ... just divide the labelled count by the permutations of the size-pattern only where sizes repeat.
- Identical into labelled: C(n + r − 1, r − 1) (stars and bars), including empty groups; with each group non-empty it becomes C(n − 1, r − 1).
- Identical into unlabelled: number of partitions of n into at most r parts — small cases are enumerated, not formulated.
- Cards and players: dealing 52 cards equally to 4 players = 52!/(13!)⁴ — labelled groups (players) of equal size, so no 4! correction; the same cards into 4 unlabelled piles would divide by 4!.
- Committee with constraints: "at least one girl" style conditions are solved by complement (total minus no-girl) — distribution logic grafted onto selection.
Counting teams: the identical-group correction
Why does 12 into three teams of 4 give 5775 and not 34650? Choose 4 of 12, then 4 of the remaining 8: C(12,4)·C(8,4)·C(4,4) = 12!/(4!·4!·4!) = 34650 counts the teams as ordered — team picked first, second, third. If the teams are genuinely indistinguishable (three groups to do the same task, no names), each unordered split was counted 3! times, so the answer is 34650/6 = 5775. The moment the question names the groups — team A, team B, team C, or three rooms, or three coaches — the division disappears and 34650 stands.
Now the identical-objects mirror: distribute 12 identical toffees among 4 children, each getting at least one. Give each child one toffee mentally, leaving 8 identical toffees for free distribution: C(8 + 3, 3) = C(11,3) = 165. Without the "at least one" constraint it would be C(15,3) = 455. The two families sit side by side in every paper: distinct objects care about who is in which group (factorials), identical objects care only about how many per group (combinations with repetition). Diagnose the two adjectives — objects identical? groups labelled? — before computing anything, because no formula rescues a wrong model.
The overcount students never see
The exam does not announce the trap; it hides it in words like "teams", "piles", "groups" versus "rows", "boxes", "prizes". Three distinct prizes among 10 students, any student eligible for any number: labelled recipients and distinct objects, so 10³ = 1000. Three identical prizes to the same 10 students: stars and bars, C(10 + 3 − 1, 3 − 1) = C(12,2) = 66 — a fifteenfold collapse caused by one adjective. Advanced layers distribution inside arrangement problems: arrange 5 boys and 3 girls in a row so no two girls sit together. The 5 boys create 6 gaps (including ends), the gaps are labelled slots, and placing the girls is choosing 3 gaps and arranging: C(6,3) × 5! × 3! = 14400. The principle stands across all of it: whenever equal-sized unnamed groups appear, hunt for the missing factorial division; whenever groups are named, delete it.
Frequently asked questions
In how many ways can 12 students form three equal unnamed teams?
12!/(4!·4!·4!·3!) = 5775; the 3! removes the ordering of the three identical-size teams.
How many ways can 12 identical toffees go to 4 children with each getting at least one?
C(11,3) = 165 by stars and bars after pre-assigning one toffee per child.
When do you divide by the factorial of group sizes?
Only when the groups are unlabelled and two or more groups have the same size; labelled groups or unequal sizes need no correction.
What is the count for distributing n distinct objects freely into r labelled boxes?
r^n, since each object independently chooses one of r boxes.
How does dealing 52 cards among 4 players differ from making 4 unnamed piles?
Players are labelled, so the count is 52!/(13!)⁴; unnamed piles divide this by 4!.