Properties of Combinations

In short

The combination \binom{n}{r} satisfies several identities that make computation faster and reveal hidden structure. The symmetry property says \binom{n}{r} = \binom{n}{n-r}. The Pascal identity says \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}. These two rules, together with properties like \binom{n}{0} = \binom{n}{n} = 1 and the Vandermonde identity, are the backbone of Pascal's triangle — an infinite triangular array where each entry is the sum of the two entries above it.

A class of 20 students needs to pick a team of 14 for a quiz competition. You sit down to compute \binom{20}{14}:

\binom{20}{14} = \frac{20!}{14!\,6!}

That is a product of enormous numbers. But notice something: choosing 14 students to be on the team is the same as choosing 6 students to be off the team. Every selection of 14 members determines a unique set of 6 non-members, and vice versa. So:

\binom{20}{14} = \binom{20}{6} = \frac{20!}{6!\,14!} = 38{,}760

The second form is far easier to compute — the numerator 20 \times 19 \times 18 \times 17 \times 16 \times 15 = 27{,}907{,}200 divided by 6! = 720. That shortcut is not a trick; it is a property of combinations, and it is one of several identities that make combinations far more manageable than their factorial definitions suggest.

The symmetry property

Symmetry of $\binom{n}{r}$

For any integers 0 \leq r \leq n:

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

Proof. Start from the definition:

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

Replace r with n - r:

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

The two expressions are identical — the factors in the denominator are the same pair, just written in the opposite order.

Why it matters. The symmetry property tells you that the two ends of any row of \binom{n}{r} values are mirrors of each other. \binom{10}{3} = \binom{10}{7}, \binom{100}{2} = \binom{100}{98}. Whenever r > n/2, replace it with n - r and save yourself a mountain of arithmetic.

Choosing $2$ objects from $\{A, B, C, D, E\}$ is the same as choosing $3$ objects to leave behind. Every selection of $2$ determines a unique complementary set of $3$, so $\binom{5}{2} = \binom{5}{3} = 10$.

Pascal's identity

For integers 1 \leq r \leq n:

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

Algebraic proof. Write out both terms on the left:

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

To add these fractions, find a common denominator. The LCD is r!\,(n-r+1)!, because r! = r \cdot (r-1)! and (n-r+1)! = (n-r+1) \cdot (n-r)!. So:

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

= \frac{n!\,\bigl[(n-r+1) + r\bigr]}{r!\,(n-r+1)!} = \frac{n!\,(n+1)}{r!\,(n+1-r)!} = \frac{(n+1)!}{r!\,(n+1-r)!} = \binom{n+1}{r}

Combinatorial proof (why this identity is true, not just algebraically valid). You have n + 1 objects. Single out one special object — call it X. Every r-element subset of the n + 1 objects either contains X or does not.

Contains X: the remaining r - 1 objects must come from the other n objects. The count is \binom{n}{r-1}.
Does not contain X: all r objects come from the other n. The count is \binom{n}{r}.

Since these two cases are exhaustive and mutually exclusive, the total is \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}.

Pascal's identity by case-splitting on a special element $X$. Subsets of size $r$ from $n+1$ objects either include $X$ (and need $r-1$ more from the remaining $n$) or exclude $X$ (and need all $r$ from the remaining $n$). The sum of these two counts is $\binom{n+1}{r}$.

Other useful properties

Here are four more identities, each with a short proof.

Property 1: Boundary values.

\binom{n}{0} = 1, \qquad \binom{n}{n} = 1

There is exactly one way to choose nothing, and exactly one way to choose everything. From the formula: \binom{n}{0} = n!/(0!\,n!) = 1 and \binom{n}{n} = n!/(n!\,0!) = 1.

Property 2: Sum of a row.

\sum_{r=0}^{n} \binom{n}{r} = 2^n

Proof. Each of the n objects is either included or excluded from a subset — 2 independent choices per object, giving 2^n subsets in total. The left side counts these same subsets by size: \binom{n}{0} subsets of size 0, plus \binom{n}{1} of size 1, and so on up to \binom{n}{n} of size n. Both sides count the same collection, so they are equal.

Alternatively, set x = 1 in the binomial theorem (1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r to get 2^n = \sum_{r=0}^{n} \binom{n}{r}.

Property 3: Alternating sum.

\sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0

Proof. Set x = -1 in the binomial theorem: (1 + (-1))^n = 0^n = 0 for n \geq 1.

This says the sum of the even-indexed entries in any row of Pascal's triangle equals the sum of the odd-indexed entries.

Property 4: Product identity (absorption).

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

Proof. The left side:

r \cdot \frac{n!}{r!\,(n-r)!} = \frac{n!}{(r-1)!\,(n-r)!}

The right side:

n \cdot \frac{(n-1)!}{(r-1)!\,(n-r)!} = \frac{n!}{(r-1)!\,(n-r)!}

They are identical. The combinatorial meaning: choosing r people from n and then marking one of the r as the leader (r choices for the leader) is the same as first picking the leader from n people (n choices) and then choosing the remaining r - 1 teammates from the other n - 1.

The absorption identity $r\binom{n}{r} = n\binom{n-1}{r-1}$ counts the same thing two ways. Path 1: pick the team first, then name a leader from it. Path 2: name the leader first, then pick the rest of the team from the remaining people.

Pascal's triangle

Pascal's identity \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} gives a recipe for building an infinite triangle of numbers row by row. Start with \binom{0}{0} = 1 at the top. Each subsequent entry is the sum of the two entries directly above it (with missing entries treated as 0).

Row 0: 1 Row 1: 1 \quad 1 Row 2: 1 \quad 2 \quad 1 Row 3: 1 \quad 3 \quad 3 \quad 1 Row 4: 1 \quad 4 \quad 6 \quad 4 \quad 1 Row 5: 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 Row 6: 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1

The entry in row n at position r (counting from 0) is \binom{n}{r}. Every property from this article is visible in the triangle:

Symmetry: each row reads the same forwards and backwards.
Boundary values: each row starts and ends with 1.
Row sum: the sum of the entries in row n is 2^n (1, 2, 4, 8, 16, 32, 64, \ldots).
Pascal's identity: every interior entry is the sum of the two entries above it.

Pascal's triangle through row $6$. The highlighted entry $6$ in row $4$ equals $\binom{4}{2}$, and it is the sum of $\binom{3}{1} = 3$ and $\binom{3}{2} = 3$ from the row above. The highlighted $20$ in row $6$ is $\binom{6}{3} = \binom{5}{2} + \binom{5}{3} = 10 + 10$. Row sums double with each new row: $1, 2, 4, 8, 16, 32, 64$.

Building rows by hand

To construct row n of Pascal's triangle without computing any factorials:

Start with 1 on the left.
Multiply the previous entry by (n - r + 1)/r to get the next entry at position r.
Stop when you reach the middle (the row is symmetric, so you can mirror it).

For row 7: start with 1. Multiply by 7/1 = 7. Multiply by 6/2 = 3, giving 21. Multiply by 5/3, giving 35. The row is 1, 7, 21, 35, 35, 21, 7, 1.

This trick comes from the ratio \binom{n}{r}/\binom{n}{r-1} = (n - r + 1)/r, which is a direct consequence of the factorial formula.

Patterns hidden in the triangle

The triangle holds more than just combination values. A few patterns worth knowing:

Column 1: the natural numbers 1, 2, 3, 4, 5, \ldots (these are \binom{n}{1}).
Column 2: the triangular numbers 1, 3, 6, 10, 15, \ldots (these are \binom{n}{2} = n(n-1)/2).
Diagonal sums: the sum along each "shallow diagonal" gives the Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, \ldots

The row sums of Pascal's triangle grow as $2^n$ — each bar is exactly twice the height of the previous one. Row $6$ sums to $64 = 2^6$, matching the identity $\sum_{r=0}^{n}\binom{n}{r} = 2^n$.

Two worked examples

Example 1: Verify that $\binom{8}{3} + \binom{8}{4} = \binom{9}{4}$ and interpret the result

Step 1. Compute \binom{8}{3}.

\binom{8}{3} = \frac{8!}{3!\,5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56

Why: the numerator is the product of the top 3 factors of 8!, and the denominator cancels the overcounting from order.

Step 2. Compute \binom{8}{4}.

\binom{8}{4} = \frac{8!}{4!\,4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = \frac{1680}{24} = 70

Why: same method, but with 4 factors in the numerator and 4! in the denominator.

Step 3. Add them.

56 + 70 = 126

Step 4. Compute \binom{9}{4} independently.

\binom{9}{4} = \frac{9!}{4!\,5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126

Why: this confirms Pascal's identity \binom{8}{3} + \binom{8}{4} = \binom{9}{4} with concrete numbers.

Step 5. Interpret.

You have 9 players trying out for a 4-member team. Single out one player — say the captain from last year. Teams that include the captain need 3 more members from the other 8: that gives \binom{8}{3} = 56 teams. Teams that exclude the captain need all 4 from the other 8: that gives \binom{8}{4} = 70 teams. Together: 56 + 70 = 126 = \binom{9}{4}. Pascal's identity is not just algebra — it counts a real split.

Result. \binom{8}{3} + \binom{8}{4} = 56 + 70 = 126 = \binom{9}{4}.

Pascal's identity at work. Splitting the $126$ four-member teams by whether they include P1 gives two non-overlapping groups: $56$ teams with P1 ($3$ more from $8$) and $70$ without ($4$ from $8$). The identity $\binom{8}{3} + \binom{8}{4} = \binom{9}{4}$ is the arithmetic version of this split.

The diagram confirms the algebra: every one of the 126 teams falls into exactly one of the two boxes, and the counts add up perfectly.

Example 2: Use Pascal's triangle to find $\binom{7}{3}$ without computing any factorials

Step 1. Write out Pascal's triangle from row 0 to row 7, using the rule "each entry is the sum of the two entries above it."

Row 0: 1. Row 1: 1, 1. Row 2: 1, 2, 1. Row 3: 1, 3, 3, 1.

Why: each row begins and ends with 1; every interior entry is the sum of the two above it. Row 3, for instance: 1, then 1+2=3, then 2+1=3, then 1.

Step 2. Continue to row 4, row 5, row 6.

Row 4: 1, 4, 6, 4, 1. Row 5: 1, 5, 10, 10, 5, 1. Row 6: 1, 6, 15, 20, 15, 6, 1.

Why: you only need addition — no multiplication, no division, no factorials.

Step 3. Build row 7.

Row 7: 1, 7, 21, 35, 35, 21, 7, 1.

The entries: 1, 1+6=7, 6+15=21, 15+20=35, 20+15=35, 15+6=21, 6+1=7, 1.

Why: the symmetry means you only need to compute the first half; the second half is a mirror.

Step 4. Read off \binom{7}{3}.

Position 3 (counting from 0) in row 7 is 35.

Result. \binom{7}{3} = 35.

Quick check: \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35. The triangle and the formula agree.

Pascal's triangle through row $7$. The highlighted entry $35$ at position $3$ in row $7$ is $\binom{7}{3}$, obtained by adding $15$ and $20$ from row $6$ — no factorial calculations needed. The symmetry $35 = 35$ at positions $3$ and $4$ reflects $\binom{7}{3} = \binom{7}{4}$.

The triangle gives you the same answer as the formula, and for small values of n it is often faster — you only need addition. For large n, the formula is more practical, but the triangle remains the single best tool for seeing how combinations relate to each other.

Common confusions

"\binom{n}{r} = \binom{n}{n+r}." The symmetry property is \binom{n}{r} = \binom{n}{n-r}, not \binom{n}{n+r}. Watch the sign: it is n minus r, not n plus r. \binom{n}{n+r} is undefined for r > 0 because you cannot choose more objects than you have.
"Pascal's identity works for r = 0." The identity \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} requires r \geq 1, because \binom{n}{-1} is not defined. At r = 0, the boundary value \binom{n+1}{0} = 1 stands on its own.
"The row sum is 2n instead of 2^n." Row 4 sums to 1 + 4 + 6 + 4 + 1 = 16 = 2^4, not 2 \times 4 = 8. The growth is exponential, not linear.
"Each entry in Pascal's triangle is the product of the two entries above it." It is the sum, not the product. The product rule would give wildly different numbers.
"\binom{n}{r} always increases as r increases." It increases up to the middle of the row and then decreases (by symmetry). For even n, the maximum is at r = n/2; for odd n, the two middle entries \binom{n}{\lfloor n/2 \rfloor} and \binom{n}{\lceil n/2 \rceil} are equal.

Going deeper

If you can state and use the symmetry rule, Pascal's identity, the row-sum formula, and build Pascal's triangle by hand, you have everything you need for combinations with restrictions and division and distribution. The rest of this section is for readers who want to see the algebraic connections and a famous pattern lurking in the triangle.

The Vandermonde identity

Suppose you have two separate groups of people — m men and n women — and you want to choose a committee of r people total. The committee could have 0 men and r women, or 1 man and r-1 women, or 2 men and r-2 women, and so on. Counting each case and summing:

\binom{m+n}{r} = \sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k}

This is the Vandermonde identity (also called the Vandermonde convolution). Setting m = n = r gives the useful special case:

\binom{2n}{n} = \sum_{k=0}^{n} \binom{n}{k}^2

The sum of squares of the entries in row n of Pascal's triangle equals the central entry of row 2n.

Hockey-stick identity

Pick any diagonal in Pascal's triangle and add consecutive entries. The sum equals the entry one step diagonally below the last one you added:

\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}

For example, \binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \binom{5}{2} = 1 + 3 + 6 + 10 = 20 = \binom{6}{3}. The pattern traces a "hockey stick" shape in the triangle — a straight shaft of entries ending in a blade that turns one step to the side. The proof is a telescoping application of Pascal's identity.

Connection to the binomial theorem

The reason combinations are also called binomial coefficients is that they are exactly the coefficients in the expansion of (a + b)^n:

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

Row n of Pascal's triangle gives the coefficients of (a+b)^n. For instance, (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4, with coefficients 1, 4, 6, 4, 1 matching row 4.

Every identity in this article has a binomial-theorem interpretation. The row sum (2^n) comes from a = b = 1. The alternating sum (0) comes from a = 1, b = -1. Pascal's identity corresponds to the recurrence that builds the polynomial expansion one degree at a time. The algebraic and the combinatorial viewpoints reinforce each other at every step.

Where this leads next

The properties of combinations unlock the harder counting problems.

Combinations with Restrictions — using the symmetry and addition rules to count selections where certain objects must (or must not) be included.
Division and Distribution — distributing objects into groups, where the formulas lean heavily on the properties proved here.
Combinations — Basics — the foundation: what \binom{n}{r} means, the formula, and the relationship to permutations.
Factorial Notation — the notation that all these identities are built on.
Algebraic Identities — the binomial theorem and related expansions, where Pascal's triangle appears as the coefficient table.