In short

A combination is a selection of r objects from n distinct objects where order does not matter. The count is \binom{n}{r} = \frac{n!}{r!\,(n-r)!}, which equals {}^nP_r / r! — the permutation count divided by the r! orderings of each selection. Two selections are the same combination if they contain the same objects, regardless of the order in which they were picked.

A school cricket team has 15 players on its roster, but only 11 can play in a match. The coach must pick 11 players. The order in which the coach announces the names does not matter — announcing "Rohit, Virat, Jasprit, ..." is the same squad as "Virat, Jasprit, Rohit, ...". What matters is which 11 names are on the list, not the sequence.

If order mattered, the count would be {}^{15}P_{11} = \frac{15!}{4!} — an enormous number. But since the squad is the same regardless of how you list the 11 names, every group of 11 players has been counted 11! times (once for each ordering of those 11 players). Dividing removes the overcounting:

\frac{{}^{15}P_{11}}{11!} = \frac{15!}{4! \times 11!} = \frac{15!}{11!\cdot 4!} = 1{,}365

The coach has 1{,}365 possible squads. This number — \frac{n!}{r!(n-r)!} — is the combination formula, and the entire article is about why it works, how to use it, and how it relates to permutations.

What is a combination?

Combination

A combination of r objects from n distinct objects is an unordered selection of r of those objects. Two combinations are the same if and only if they contain the same elements.

The number of such combinations is denoted \binom{n}{r} (read "n choose r"), also written {}^nC_r or C(n, r).

The word "unordered" is the key. In a permutation, the selection {A, B, C} and the selection {C, A, B} are counted as two different arrangements. In a combination, they are the same selection — the same three objects, just listed differently.

Permutations versus combinations: choosing 3 from 4 objectsLeft side shows all 24 permutations of 3 objects from A B C D, grouped in fours. Right side shows the 4 combinations, each listed once. Each group of permutations on the left maps to one combination on the right, with a factor of 3 factorial equals 6 connecting them. Permutations: ⁴P₃ = 24 Combinations: C(4,3) = 4 ABC ACB BAC BCA CAB CBA {A,B,C} ← 6 perms ABD ADB BAD BDA DAB DBA {A,B,D} ACD + 5 more {A,C,D} BCD + 5 more {B,C,D} 24 permutations 4 combinations
Choosing $3$ from $\{A, B, C, D\}$: every group of $3! = 6$ permutations on the left (differing only in order) collapses into a single combination on the right. The ratio is $24/6 = 4 = \binom{4}{3}$.

Deriving the \binom{n}{r} formula

The derivation has a single core idea: a permutation = a combination + an arrangement of that combination.

Step 1. Start with the permutation count. The number of ways to pick r objects from n and arrange them in order is:

{}^nP_r = \frac{n!}{(n-r)!}

Step 2. Every permutation can be broken into two stages:

By the multiplication principle:

{}^nP_r = \binom{n}{r} \times r!

Step 3. Solve for \binom{n}{r}:

\binom{n}{r} = \frac{{}^nP_r}{r!} = \frac{n!}{r!\,(n-r)!}
Derivation: permutation equals combination times r factorialA flow diagram. A box labelled nPr with an arrow splits into two boxes: one labelled C(n,r) for selection and one labelled r factorial for arrangement. The equation nPr equals C(n,r) times r factorial is shown below, rearranged to give the combination formula. ⁿPᵣ = n!/(n−r)! Select r objects C(n, r) = ? × Arrange r objects r! C(n, r) = ⁿPᵣ / r! = n! / [r!(n−r)!]
A permutation of $r$ from $n$ is a selection (combination) followed by an arrangement. Since the arrangement of $r$ objects has $r!$ orderings, dividing the permutation count by $r!$ isolates the combination count.

The combination formula

\binom{n}{r} = \frac{n!}{r!\,(n-r)!} \qquad (0 \leq r \leq n)

Equivalently: \binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!} (the product of r descending integers from n, divided by r!).

Special cases:

Symmetry of combinations: choosing r to include equals choosing n minus r to excludeA row of 5 circles representing 5 objects. Three are filled (selected) and two are empty (excluded). An arrow labelled C(5,3) points to the filled circles. Another arrow labelled C(5,2) points to the empty circles. Text says C(5,3) equals C(5,2) equals 10. 5 objects: choose 3 to include = choose 2 to exclude A B C D E C(5, 3) = 10 selected C(5, 2) = 10 excluded C(5,3) = C(5,2) = 10 — same count
Choosing $3$ objects to include from $5$ is the same decision as choosing $2$ objects to exclude. The symmetry $\binom{5}{3} = \binom{5}{2} = 10$ follows from the formula: both give $\frac{5!}{3!\cdot 2!}$.

The difference from permutations

The distinction between permutations and combinations comes down to one question: does the order of selection matter?

Permutation Combination
Order Matters Does not matter
Formula {}^nP_r = \frac{n!}{(n-r)!} \binom{n}{r} = \frac{n!}{r!(n-r)!}
Relationship — \binom{n}{r} = {}^nP_r / r!
Example Assigning 3 prizes (1st, 2nd, 3rd) to 10 runners Choosing 3 runners for a relay team

The combination count is always \leq the permutation count, and the ratio is exactly r!:

{}^nP_r = r! \times \binom{n}{r}

For r = 0 or r = 1, the two counts are equal (since 0! = 1! = 1). For r \geq 2, the permutation count is strictly larger.

Comparing nPr and C(n,r) for n equals 5A bar chart with r from 0 to 5 on the horizontal axis. For each r, two bars are shown: a taller bar for 5Pr and a shorter bar for C(5,r). The values are: r=0 both 1; r=1 both 5; r=2, 20 and 10; r=3, 60 and 10; r=4, 120 and 5; r=5, 120 and 1. The ratio between the bars grows as r increases. n = 5: Permutations vs Combinations r=0 r=1 r=2 r=3 r=4 r=5 1 5 20 10 60 10 120 5 120 1 ⁿPᵣ C(n,r)
For $n = 5$: the permutation count (dark bars) grows rapidly with $r$, while the combination count (accent bars) rises to a peak at $r = 2$ or $3$ and then falls symmetrically. The ratio ${}^5P_r / \binom{5}{r} = r!$ grows as $r$ increases.

How to decide which formula to use. Read the problem and ask: "If I rearrange the selected objects, do I get a different outcome?"

Some more examples:

Computing \binom{n}{r} efficiently

The formula \frac{n!}{r!(n-r)!} involves factorials that can be very large. A practical computation trick: use the "descending product" form and cancel step by step.

\binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}

This has r factors in the numerator and r factors in the denominator. Always choose the smaller of r and n - r (using the symmetry \binom{n}{r} = \binom{n}{n-r}) to minimise computation.

For example, \binom{10}{7} = \binom{10}{3} (since 10 - 7 = 3 < 7):

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120

Three multiplications and one division, instead of computing 10! = 3{,}628{,}800 and dividing by 7! \times 3!.

Efficient computation of C(10,3)The formula C(10,3) equals 10 times 9 times 8 in the numerator, divided by 3 times 2 times 1 in the denominator, giving 720 over 6 equals 120. The step of using symmetry to replace C(10,7) with C(10,3) is shown. C(10, 7) = C(10, 3) ← use smaller index numerator: 10 × 9 × 8 = 720 denominator: 3 × 2 × 1 = 6 C(10, 3) = 720 / 6 = 120
Instead of computing $10!$, $7!$, and $3!$ separately, use the descending-product form with the smaller index: $3$ multiplications in the numerator and $1$ in the denominator. The symmetry $\binom{10}{7} = \binom{10}{3}$ halves the work.

Two worked examples

Example 1: A committee of $4$ is to be formed from $6$ men and $4$ women. How many committees are possible if there are no restrictions?

Step 1. Identify n and r.

There are 6 + 4 = 10 people in total, and you are selecting 4 for the committee.

Why: the committee members are equals — no president, no secretary — so the order of selection does not matter. This is a combination problem.

Step 2. Apply the formula.

\binom{10}{4} = \frac{10!}{4!\cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5{,}040}{24} = 210

Why: 4 descending factors from 10 give the numerator, and 4! in the denominator removes the 4! orderings of each committee. Using the descending-product form avoids computing large factorials.

Step 3. Verify the permutation-to-combination ratio.

{}^{10}P_4 = 10 \times 9 \times 8 \times 7 = 5{,}040. Dividing by 4! = 24 gives 5{,}040/24 = 210. The ratio is 4! = 24, meaning each committee of 4 people corresponds to 24 ordered selections. This confirms the formula.

Why: if the committee had 4 distinct roles (chair, secretary, treasurer, member), the count would be 5{,}040 — a permutation. Removing the roles (making it unordered) divides by 4! = 24.

Step 4. Sanity check with a smaller instance.

Choosing 2 from 4 people: \binom{4}{2} = 6. Listing: {A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C,D} — six committees. The formula works.

Result. \binom{10}{4} = 210 committees.

Selecting 4 from 10 people for a committeeTen circles in a row, representing 10 people. Four of them (persons 2, 5, 7, 9) are highlighted in red. Text says: this is one of C(10,4) equals 210 possible committees. Below, the note: swapping the order of the 4 selected people does not create a new committee. 10 people — select 4 (order does not matter) 1 2 3 4 5 6 7 8 9 10 One committee: {2, 5, 7, 9} {2,5,7,9} = {9,7,5,2} = {5,2,9,7} — same committee C(10, 4) = 210 committees each committee has 4! = 24 orderings, all counted as one
One of the $210$ committees: persons $2, 5, 7, 9$. Rearranging the four selected people produces the same committee — this is what makes it a combination, not a permutation.

Each committee is a set of 4 people. The committee \{2, 5, 7, 9\} is the same regardless of whether you picked person 2 first or person 9 first. The 210 committees represent 210 distinct groups, and each has 4! = 24 internal orderings, giving 210 \times 24 = 5{,}040 = {}^{10}P_4 permutations in total.

Example 2: A bag contains $5$ red balls and $3$ blue balls (each ball is distinguishable). In how many ways can you draw $4$ balls such that exactly $2$ are red?

Step 1. Break the selection into two independent stages.

Stage 1: choose 2 red balls from 5. Stage 2: choose 2 blue balls from 3. The stages are independent — the choice of red balls does not affect the choice of blue balls.

Why: "exactly 2 red" forces the remaining 2 to be blue (4 - 2 = 2). Each stage is an unordered selection (you are drawing balls, not arranging them), so each stage is a combination.

Step 2. Count each stage.

\text{Red:} \quad \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10
\text{Blue:} \quad \binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3

Why: \binom{5}{2} selects 2 of the 5 red balls; \binom{3}{2} selects 2 of the 3 blue balls. Both are direct applications of the combination formula.

Step 3. Combine by the multiplication principle.

\binom{5}{2} \times \binom{3}{2} = 10 \times 3 = 30

Why: for each of the 10 ways to pick the red pair, there are 3 ways to pick the blue pair, independently. The multiplication principle gives 10 \times 3 = 30 total draws.

Step 4. Sanity check.

The total ways to draw 4 balls from 8 (without the colour restriction) is \binom{8}{4} = 70. The restricted count (30) is less than 70, as expected. The fraction 30/70 \approx 43\% is reasonable — roughly 4 out of 10 draws have exactly 2 red balls.

Result. There are 30 ways to draw 4 balls with exactly 2 red.

Choosing 2 red from 5 and 2 blue from 3Two groups of circles. Left group has 5 red circles labelled R1 through R5, with R2 and R4 highlighted. Right group has 3 blue circles labelled B1 through B3, with B1 and B3 highlighted. Below, C(5,2) equals 10 times C(3,2) equals 3, giving 30 total. 5 red balls 3 blue balls R₁ R₂ R₃ R₄ R₅ B₁ B₂ B₃ C(5, 2) = 10 × C(3, 2) = 3 = 30 ways to draw exactly 2 red, 2 blue
Choose $2$ red balls (from $5$) and $2$ blue balls (from $3$) independently. The multiplication principle gives $\binom{5}{2} \times \binom{3}{2} = 10 \times 3 = 30$.

This "stage-by-stage" approach extends naturally. If you needed exactly 3 red and 1 blue, the count would be \binom{5}{3} \times \binom{3}{1} = 10 \times 3 = 30. If you needed at least 1 blue, you could use complementary counting: \binom{8}{4} - \binom{5}{4} = 70 - 5 = 65 (subtract the all-red draws).

Common confusions

Going deeper

If you can compute \binom{n}{r}, distinguish combinations from permutations, and apply the formula to selection problems, you are ready for properties of combinations and combinations with restrictions. The rest of this section connects the combination formula to deeper mathematical ideas.

Pascal's triangle and the recursion

The combinations \binom{n}{r} satisfy a beautiful recursion:

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

Why this is true. Consider the n-th object. Every combination of r objects from n either includes object n or excludes it. If it includes object n, the remaining r - 1 objects are chosen from the other n - 1: that gives \binom{n-1}{r-1} combinations. If it excludes object n, all r objects are chosen from the other n - 1: that gives \binom{n-1}{r}. The two cases are disjoint and exhaustive, so the counts add.

This recursion generates Pascal's triangle — the triangular array where each entry is the sum of the two entries above it:

\begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 1 & & 2 & & 1 & & & \\ & & 1 & & 3 & & 3 & & 1 & & \\ & 1 & & 4 & & 6 & & 4 & & 1 & \\ 1 & & 5 & & 10 & & 10 & & 5 & & 1 \\ \end{array}

Row n (starting from n = 0) contains the values \binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}.

The binomial theorem

The combinations \binom{n}{r} are also called binomial coefficients because they appear as coefficients in the expansion of a binomial power:

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r

For n = 3: (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3, with coefficients 1, 3, 3, 1 = \binom{3}{0}, \binom{3}{1}, \binom{3}{2}, \binom{3}{3}.

The connection is combinatorial: expanding (a+b)^n means multiplying n factors of (a+b). Each term in the expansion corresponds to choosing a from some factors and b from the rest. The coefficient of a^{n-r}b^r counts the number of ways to choose r of the n factors to contribute b — that is \binom{n}{r}.

Combinations with repetition

What if you can repeat selections — choosing r items from n types where the same type can be chosen more than once (like choosing 3 scoops from 8 flavours, where you can pick the same flavour multiple times)? The count is:

\binom{n + r - 1}{r} = \frac{(n+r-1)!}{r!(n-1)!}

This is the "stars and bars" formula. It counts the number of ways to distribute r identical items into n distinct bins. The derivation uses a bijection between such distributions and sequences of r stars and n - 1 bars — a topic for a future article.

Where this leads next