Combinations: Formula, Types & Examples

Types of Combinations

In combinatorics, combinations represent ways to select or distribute items without considering order, in contrast to permutations where order matters. Building on our earlier comparison between combinations and permutations, we now distinguish between different types of combinations using key questions: whether the objects are distinct or identical, whether we are selecting a subset or distributing all objects, and whether the containers (cells) are labeled or can remain empty.

Combination Scenarios

Scenario	Objects Distinct Or Identical?	Select Subset Or Distribute?	Cells Labeled Or Unlabeled?	Empty Cells Allowed?	Formula	Example
Simple Combination	Distinct	Select	N/A	N/A	$\binom{n}{r}$	Choose 3 fruits out of 10 different types
Partition into Groups	Distinct	Distribute	Labeled	Typically no	$\frac{n!}{n_1!n_2!\dots n_k!}$ (divide by $k!$ when equal-size groups are unlabeled)	Split 12 players into 3 named teams of 4
Weak Composition	Identical	Distribute	Labeled	Yes	$\binom{n+r-1}{r-1}$	Distribute 7 identical candies to 3 children (some may get none)
Strong Composition	Identical	Distribute	Labeled	No	$\binom{n-1}{r-1}$	Distribute 7 identical candies to 3 children (each gets at least 1)
Distribution into Cells	Distinct	Distribute	Labeled	Yes	$r^n$	Assign 5 employees to 3 different projects

Conclusion:
Understanding these different combination scenarios is essential because each situation leads to a different counting formula. Recognizing the correct type of combination ensures accurate problem-solving in probability, statistics, and many applied fields.

Simple Combinations

Counting ordered picks already works, so a new method has to explain itself.

Choose

2

of the

3

balls numbered

1

,

2

,

3

. Pick one ball —

3

choices — then another —

2

choices left — and the multiplication principle gives

3 \times 2 = 6

. That count is correct and quick. It is the number of partial permutations,

P(3, 2) = 6

.

All

6

ordered picks of

2

from

3

, each with a first and a second position. The same state is on the partial permutation visualizer.

When order does not matter, the ordered count counts every group twice.

A team, a hand of cards, a set of books: what matters is who is in, not who was picked first. Sort the

6

picks by the group they end up as.

the set	ordered picks that give it	times counted
{1, 2}	1 then 2, 2 then 1	2
{1, 3}	1 then 3, 3 then 1	2
{2, 3}	2 then 3, 3 then 2	2
3 sets	6 picks	6 ÷ 2 = 3

Each group of

2

appears once for every way of ordering it —

2! = 2

times. Half of the

6

is repetition. With

5

balls taken

3

at a time the waste is worse:

P(5, 3) = 60

picks for only

10

groups, every group counted

3! = 6

times.

The

3

groups that are actually different. Each one stands for

2

of the ordered picks above. The same state is on the combination visualizer.

The definition

A simple combination is a selection of

r

items from

n

distinct items in which order does not matter and no item is used twice. The number of them is

Combination

\binom{n}{r} = \frac{n!}{r! \, (n-r)!}.

Learn more about this formula: Combination →

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A combination is a set, and the formula counts sets.

Each combination is an unordered group —

\{1, 2\}

and

\{2, 1\}

are the same one — and

\binom{n}{r}

is a whole number: how many such groups there are.

Simple Combination works when selecting items from a collection where order doesn't matter and no repetitions are allowed. Classic examples include choosing team members from a group or selecting lottery numbers.

Property	Simple Combination
Order matters	✗
Objects distinct or identical?	Distinct
Select subset or distribute?	Select subset
Cells labeled or unlabeled?	N/A
Empty cells allowed?	N/A

Simple Combination — Examples:
Choosing committee members from a club, selecting lottery numbers, picking a set of books to borrow from a library, forming a group of students for a project, selecting ingredients for a recipe from available options.

Simple Combinations Examples

Two facts come with the formula for free.

First, it is the ordered count with the waste divided out:

\binom{n}{r} = \frac{P(n, r)}{r!}

— for our balls,

\frac{6}{2} = 3

. Second, choosing which

r

to take is the same as choosing which

n - r

to leave, so

\binom{n}{r} = \binom{n}{n-r}

.

take r of 3	leave 3 − r	C(3, r)
0	3	1
1	2	3
2	1	3
3	0	1

Common notations:
•

\binom{n}{r}

- "n choose r" (most common in modern texts)
•

C(n,r)

- C for "combinations"
•

C_n^r

- alternative C notation
•

_nC_r

- subscript/superscript form

The

\binom{n}{r}

notation is the most widely used in modern mathematics, but you'll often see

C(n,r)

in introductory texts and

_nC_r

on calculators.

Learn more about combinatorial notations and symbols here →

Everything rests on one fact: every group is counted the same number of times.

Dividing by

r!

is honest only because each group of

r

distinct items has exactly

r!

orderings. That is what the conditions buy:

Here

n

is the total number of distinct items and

r

is the number of items to select; order doesn't matter and repetition isn't allowed.

Break them and the single division fails. If some items are identical, or an item may be chosen more than once, different groups have different numbers of orderings, and the counts on the rest of this page take over.

The sections that follow handle what simple combinations cannot: splitting items into several labelled groups, weak and strong compositions, and distributing items into cells.

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Partition into Groups

Partition into Groups applies when dividing distinct items into labeled (distinguishable) groups of given sizes, where only which items share a group matters, not the order within groups. Classic examples include dividing students into teams or splitting items into categories.

Property	Partition into Groups
Order matters	✗
Objects distinct or identical?	Distinct
Select subset or distribute?	Distribute
Cells labeled or unlabeled?	Labeled
Empty cells allowed?	Typically no

Partition into Groups — Examples:
Dividing students into study groups, splitting employees into project teams, grouping tasks into phases of a project, dividing guests into tables at an event, forming clusters of data points in analysis.

Fact: A perfect bridge deal is hardly ever luck

The partition formula gives

\frac{52!}{(13!)^4} \approx 5.36 \times 10^{28}

bridge deals, and only

4! = 24

of them hand each seat a complete suit, a chance of about

4.5 \times 10^{-28}

. A reported deal of four whole suits is far more likely a new deck, sorted by suit, that was barely shuffled.

A game of bridge starts with exactly this kind of split. The

52

cards are dealt into four hands of

13

, and the hands are labeled by the seats North, East, South and West, so the same cards in a different seat make a different deal. The order in which the cards arrive in a hand does not matter, only which

13

cards it holds. In the formula below,

n = 52

,

k = 4

and every group size is

13

.

Notation:
Partition into Groups doesn't have a standard symbolic notation like

\binom{n}{r}

for simple combinations.

Partitions are typically just described in words or with set notation like:

"Partition of set S into k parts"
{A₁, A₂, ..., Aₖ} where A₁ ∪ A₂ ∪ ... ∪ Aₖ = S

Unlike other combinations, permutations, etc., partitions don't have a universally recognized compact notation symbol.

Learn more about combinatorial notations and symbols here →

Formula:
The total number of partitions into groups is

Partition into Groups

\frac{n!}{n_1! \, n_2! \cdots n_k!}, \quad n_1 + n_2 + \cdots + n_k = n.

Learn more about this formula: Partition into Groups →

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Pitfall: Forgetting k! for unnamed groups

Splitting

6

people into

3

pairs with no names as

\frac{6!}{2! \, 2! \, 2!} = 90

counts each split once for every one of the

3! = 6

ways to name the pairs. The number of splits is

\frac{90}{3!} = 15

.

Here

n

is the total number of distinct items to partition into

k

groups of sizes

n_1, n_2, \ldots, n_k

respectively, where the groups are labeled (distinguishable) and the order within each group does not matter.

The groups count as labeled when each one has its own name or role, such as Team A and Team B, or a morning shift and an evening shift. Exchanging the members of two such groups then gives a different split. When the groups are told apart only by who is in them, as with tables at an event, exchanging two whole groups of the same size leaves the split as it was.

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Only who shares a group counts, so the orders inside each labeled group are divided out of

n!

.

Weak Composition

Weak Composition occurs when distributing identical units into labeled containers where some containers can remain empty and only the count in each container matters. Classic examples include distributing identical coins into numbered boxes or allocating identical resources to different departments.

Property	Weak Composition
Order matters	✗
Objects distinct or identical?	Identical
Select subset or distribute?	Distribute
Cells labeled or unlabeled?	Labeled
Empty cells allowed?	✓

Weak Composition — Examples:
Distributing identical coins into numbered boxes, allocating identical resources to different departments, assigning identical tasks across several workers. Since this type of experiment is highly specific — focusing only on counts in labeled containers, even allowing some to remain empty — it is largely self‑explanatory and not as easy to find in common real‑world situations.

Notation:
The notation for Weak Composition is:

\left(\binom{r}{n}\right)

or

\left(\left(\binom{r}{n}\right)\right)

This represents the number of ways to place n identical objects into r labeled bins, where empty bins are allowed. It's also written as

\binom{n+r-1}{r-1}

but the double parentheses notation specifically indicates weak compositions.

Learn more about combinatorial notations and symbols here →

Formula:
The total number of weak compositions is

Weak Composition

\left(\!\!\binom{r}{n}\!\!\right) = \binom{n + r - 1}{r - 1}.

Learn more about this formula: Weak Composition →

Browse all combinatorics formulas →

Pitfall: Swapping items and containers

For

7

identical candies and

3

children, putting

n = 3

and

r = 7

into the formula gives

\binom{9}{6} = 84

. With

n = 7

candies and

r = 3

children, the count is

\binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} = 36

.

Here

n

is the total number of identical items to distribute into

r

labeled containers, where empty containers are allowed and only the count in each container matters.

The two numbers in the formula do different jobs. The

n

items are the units being shared out, and since they are identical, a distribution is fixed by how many units each container receives. Picture the items as a row of

n

stars split into

r

containers by

r - 1

bars: the

n + r - 1

symbols in the row give the top of the binomial coefficient, and the

r - 1

bars give the bottom.

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Identical items in labeled containers become a row of stars and bars, so the count is a choice of where the bars go.

Strong Composition

Strong Composition works when distributing identical units into labeled containers where every container must receive at least one unit and only the count in each container matters. Classic examples include distributing identical items to departments where each department must get something or allocating identical resources ensuring no group is left empty.

Property	Strong Composition
Order matters	✗
Objects distinct or identical?	Identical
Select subset or distribute?	Distribute
Cells labeled or unlabeled?	Labeled
Empty cells allowed?	✗

Strong Composition — Examples:
Distributing identical items to departments where each department must receive at least one, allocating identical resources among teams ensuring none are left empty, dividing identical tasks across several workers with no one unassigned. Like weak composition, this type of experiment is quite specific — it focuses on counts in labeled containers with no empties allowed — making it largely self‑explanatory and less common in everyday situations.

Fact: A whole number

n

has

2^{n-1}

ordered sums

Counted over every number of parts, the ways to write

n

as an ordered sum of positive whole numbers total

2^{n-1}

: line up

n

stars and decide for each of the

n - 1

gaps whether a bar goes in. The number

4

has

8

of them, from

4

itself to

1 + 1 + 1 + 1

.

A strong composition can also be read as a way of writing

n

as an ordered sum of positive whole numbers, one summand for each container. With

n = 4

and

r = 2

the sums are

1 + 3

,

2 + 2

and

3 + 1

, and the formula below agrees:

\binom{3}{1} = 3

. Since

1 + 3

and

3 + 1

fill the two containers differently, they count as different sums. Letting

r

run through

1, 2, 3

and

4

lists every such sum for

n = 4

.

Notation:
There is no standard symbol for Strong Composition; the count is written directly as:

\binom{n-1}{r-1}

This represents the number of ways to place n identical objects into r labeled bins where each bin must contain at least one object.

Learn more about combinatorial notations and symbols here →

Formula:
The total number of strong compositions is

Strong Composition

\binom{n - 1}{r - 1}.

Learn more about this formula: Strong Composition →

Browse all combinatorics formulas →

Pitfall: Allowing empty containers

Counting

7

identical candies for

3

children as

\binom{7 + 3 - 1}{3 - 1} = 36

includes splits like

7 + 0 + 0

, where a child gets nothing. With at least one candy each, the count is

\binom{7 - 1}{3 - 1} = \binom{6}{2} = 15

.

Here

n

is the total number of identical items to distribute into

r

labeled containers, where each container must receive at least one item and only the count in each container matters.

The requirement changes what is counted, not the items or the containers. One way to see the link with weak composition is to hand one unit to every container first; the remaining

n - r

units can then be shared out freely, with empty containers allowed. The condition also needs

n \geq r

, since with fewer units than containers some container has to stay empty.

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Distribution into Cells

Distribution into Cells works when assigning each distinct item to a specific labeled container, creating a mapping that shows which container each item goes to. Classic examples include assigning students to different classrooms or placing different files into labeled folders.

Property	Distribution into Cells
Order matters	✗
Objects distinct or identical?	Distinct
Select subset or distribute?	Distribute
Cells labeled or unlabeled?	Labeled
Empty cells allowed?	✓

Distribution into Cells — Examples:
Assigning students to classrooms, placing different files into labeled folders, allocating distinct products to specific storage bins, or assigning employees to designated offices. This setup focuses on mapping each unique item to a labeled container, making the outcome a mapping of items to containers, where several items may share a container.

Notation:

Distribution into Cells doesn't have a standard symbolic notation like

\binom{n}{r}

for simple combinations. It's typically just described as "functions from set A to set B" or "mappings" in mathematical notation, but there's no compact symbol specifically for this concept like there is for combinations or permutations.

Learn more about combinatorial notations and symbols here →

Formula:
The total number of distributions into cells is

Distribution into Cells

r^n.

Learn more about this formula: Distribution into Cells →

Browse all combinatorics formulas →

Pitfall: Swapping base and exponent

Assigning

5

employees to

3

projects as

5^3 = 125

treats each project as picking one employee. Each employee picks one of

3

projects, so the count is

3^5 = 243

.

Here

n

is the total number of distinct items to assign and

r

is the number of labeled containers, where each item can go into any container and containers can be empty.

The count comes from the multiplication rule applied item by item. The first item has

r

containers to go to, the second item again has

r

, and so on, because a container can take any number of items. With

n

items this is a product of

n

equal factors

r \cdot r \cdots r

, one factor for each item.

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Common Pitfalls

Four mistakes come up again and again when a counting formula from this page is put to work. Each is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Forgetting k! for unnamed groups

The formula

\frac{n!}{n_1! \, n_2! \cdots n_k!}

counts splits into labeled groups. It is used for groups that have no names, so each split is counted once for every way of attaching the names to groups of the same size.

Wrong working:

6

people are to be split into

3

pairs, with no names for the pairs. The count is given as

\frac{6!}{2! \, 2! \, 2!} = \frac{720}{8} = 90

.

Right working: the

90

counts named pairs, Pair 1, Pair 2 and Pair 3. Any one split into pairs can be named in

3! = 6

ways, so the number of splits is

\frac{90}{6} = 15

. Only groups of equal size can be exchanged like this: if all the sizes are different, the sizes already tell the groups apart and no division is needed.

Why it happens: the formula looks complete, and problems rarely say whether the groups have names, so it is easy to miss that the formula assumes they do.

How to check: count directly. Person 1 has

5

possible partners; of the

4

people left, the first has

3

possible partners; the last

2

form the third pair. That gives

5 \cdot 3 \cdot 1 = 15

.

Swapping items and containers

In

\binom{n + r - 1}{r - 1}

the number of items

n

and the number of containers

r

do different jobs, and the two are put into the formula the wrong way round.

Wrong working:

7

identical candies are shared among

3

children. With

n = 3

and

r = 7

, the count is given as

\binom{3 + 7 - 1}{7 - 1} = \binom{9}{6} = 84

.

Right working: the candies are the items and the children are the containers, so

n = 7

and

r = 3

:

\binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} = 36

.

Why it happens: the top index

n + r - 1

is the same either way, so the swap only shows in the bottom index. The notation

\left(\!\!\binom{r}{n}\!\!\right)

also writes the number of containers on top, the opposite of

\binom{n}{r}

.

How to check: try a case small enough to list.

2

candies for

3

children:

2 + 0 + 0

in

3

orders and

1 + 1 + 0

in

3

orders,

6

in all. The correct substitution gives

\binom{4}{2} = 6

; the swapped one gives

\binom{4}{1} = 4

.

Allowing empty containers

Every container must receive at least one unit, but the weak composition formula is used, and it also counts the distributions that leave some containers empty.

Wrong working:

7

identical candies for

3

children, each child getting at least one. The count is given as

\binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} = 36

.

Right working:

\binom{7 - 1}{3 - 1} = \binom{6}{2} = 15

. The same number comes from giving each child one candy first and sharing the remaining

4

freely:

\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15

.

Why it happens: the two problems differ in a single condition, and the weak formula is usually the one learned first.

How to check: the difference

36 - 15 = 21

must be the number of distributions with an empty child. Exactly one child gets nothing in

3 \cdot 6 = 18

ways (choose that child, then split

7

between the other two with at least one each), and two children get nothing in

3

ways. Indeed

18 + 3 = 21

.

Swapping base and exponent

The count

r^n

for

n

distinct items in

r

labeled containers is written as

n^r

.

Wrong working:

5

employees are assigned to

3

projects, and the count is given as

5^3 = 125

.

Right working: each of the

5

employees chooses one of the

3

projects, so the count is

3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^5 = 243

.

Why it happens: "

5

employees to

3

projects" names the numbers in the order base, then exponent. The count

5^3

answers a different question, in which each project picks one employee and one employee may be picked by several projects.

How to check: try

2

employees and

3

projects. Each employee has

3

choices, giving

9

assignments, which is

3^2

, not

2^3 = 8

. The exponent is always the number of choices made, one for each item.

Combinations FAQ

What is the difference between a permutation and a combination?

+

Permutations count arrangements where order matters; combinations count selections where it does not. Choosing three people for a podium with ranked places is a permutation, while choosing three for a committee is a combination. The combination count is always smaller, since every selection corresponds to several orderings of the same items.Read more →

Why does the combination formula divide by r factorial?

+

Because the permutation count treats each ordering of the same chosen items as separate, and there are r factorial ways to order r items. Dividing by that number collapses every ordering of a given selection into one. The division is exactly what converts a count of arrangements into a count of selections. Try values on the combinations calculator.Read more →

What is the difference between weak and strong composition?

+

A weak composition allows parts equal to zero, so some recipients may receive nothing. A strong composition requires every part to be at least one, so everyone receives something. The distinction changes the formula, since the strong case is handled by giving each recipient one item first and distributing the remainder weakly.Read more →

How many ways can you split a group into teams?

+

Divide the total arrangements by the factorial of each team size, and divide again by the factorial of the number of teams if the teams are unlabelled and all the same size. That second division catches the case where swapping two whole teams produces the same split, which is the step most often forgotten. See the partition calculator.Read more →

Is nCr the same thing as the binomial coefficient?

+

Yes. The two notations name the same number, counted differently: nCr counts selections of r items from n, while the binomial coefficient counts the coefficient of a term in an expansion. That they coincide is the reason counting and algebra connect through the binomial theorem.Read more →

What are the odds of a perfect bridge deal?

+

About

4.5 \times 10^{-28}

with a fair shuffle. The partition formula gives

\frac{52!}{(13!)^4} \approx 5.36 \times 10^{28}

ways to deal four labeled hands of

13

, and only

4! = 24

of them hand each seat a complete suit. A reported deal of four whole suits is far more likely a new deck, sorted by suit, that was barely shuffled.Read more →

What is the stars and bars method?

+

A way to count weak compositions. Line up the

n

identical items as stars and split them among

r

containers with

r - 1

bars. Each arrangement of the

n + r - 1

symbols is one distribution, so the count is the number of ways to place the bars,

\binom{n+r-1}{r-1}

. For

7

identical candies and

3

children that is

\binom{9}{2} = 36

.Read more →

How many ways can a number be written as an ordered sum of positive integers?

+

2^{n-1}

, counting every number of parts. Line up

n

stars and decide for each of the

n - 1

gaps whether a bar goes in; each set of choices gives one ordered sum. The number

4

has

8

:

4

,

1 + 3

,

3 + 1

,

2 + 2

,

1 + 1 + 2

,

1 + 2 + 1

,

2 + 1 + 1

and

1 + 1 + 1 + 1

. Adding the strong composition counts

\binom{n-1}{r-1}

over all

r

gives the same total.Read more →

Combinations

Combinations: Selecting and Grouping without Order

Types of Combinations

Combination Scenarios

Simple Combinations

Simple Combinations Examples

Partition into Groups

Weak Composition

Strong Composition

Distribution into Cells

Common Pitfalls

Combinations FAQ

What is the difference between a permutation and a combination?

Why does the combination formula divide by r factorial?

What is the difference between weak and strong composition?

How many ways can you split a group into teams?

Is nCr the same thing as the binomial coefficient?

What are the odds of a perfect bridge deal?

What is the stars and bars method?

How many ways can a number be written as an ordered sum of positive integers?