In short

The binomial coefficients \binom{n}{r} are the coefficients in the expansion of (a + b)^n. Beyond counting combinations, they satisfy a web of identities: the row sum \sum \binom{n}{r} = 2^n, the alternating sum \sum (-1)^r \binom{n}{r} = 0, the symmetry \binom{n}{r} = \binom{n}{n-r}, Pascal's identity, and the Vandermonde identity. Each of these has both an algebraic proof (plug specific values into the binomial theorem) and a combinatorial proof (count the same set in two ways).

Take the expansion of (1 + 1)^5 = 2^5 = 32. From the binomial theorem:

(1 + 1)^5 = \binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 1 + 5 + 10 + 10 + 5 + 1

Add those up: 1 + 5 + 10 + 10 + 5 + 1 = 32. So the six binomial coefficients in row 5 of Pascal's triangle sum to exactly 2^5. Is this a coincidence?

Now try (1 - 1)^5 = 0^5 = 0:

(1 - 1)^5 = \binom{5}{0} - \binom{5}{1} + \binom{5}{2} - \binom{5}{3} + \binom{5}{4} - \binom{5}{5} = 1 - 5 + 10 - 10 + 5 - 1 = 0

The same six numbers, with alternating signs, cancel to zero. These are not coincidences — they are identities that hold for every n, and the binomial theorem proves them in one line. This article develops these identities and several more.

Properties of \binom{n}{r}

Before the main identities, recall the core properties from combinations and properties of combinations:

Symmetry. \binom{n}{r} = \binom{n}{n-r}. Choosing r items to include is the same as choosing n - r items to exclude.

Pascal's identity. \binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}. Each entry in Pascal's triangle is the sum of the two entries above it.

Boundary values. \binom{n}{0} = \binom{n}{n} = 1. There is one way to choose nothing and one way to choose everything.

The product form. \binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}, with r factors in both numerator and denominator.

Summary of basic properties of binomial coefficientsFour boxed properties listed vertically. Symmetry: C(n,r) = C(n,n-r). Pascal's identity: C(n,r) = C(n-1,r-1) + C(n-1,r). Boundary: C(n,0) = C(n,n) = 1. Product form: C(n,r) = n descending r divided by r factorial. Core properties of C(n, r) Symmetry: C(n, r) = C(n, n − r) Pascal: C(n, r) = C(n−1, r−1) + C(n−1, r) Boundary: C(n, 0) = C(n, n) = 1 Product: C(n, r) = n·(n−1)···(n−r+1) / r! these are the building blocks for all the identities below
The four basic properties of $\binom{n}{r}$. Every identity in this article can be derived from these, together with the binomial theorem.

Identity 1: Sum of binomial coefficients

Sum of binomial coefficients

For every positive integer n:

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

That is, \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n.

Algebraic proof. Start from the binomial theorem:

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

Set a = 1 and b = 1:

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

The left side is 2^n. Done.

Combinatorial proof. The left side counts subsets of an n-element set, grouped by size: \binom{n}{r} subsets have exactly r elements. The right side counts all subsets directly: each of the n elements is either in or out, giving 2^n subsets. Both sides count the same collection.

Sum of row 4 of Pascal's triangle equals 2 to the 4Row 4 of Pascal's triangle: 1 + 4 + 6 + 4 + 1 = 16 = 2 to the 4. Below, a set of 4 elements is shown with 16 subsets grouped by size: 1 empty, 4 singletons, 6 pairs, 4 triples, 1 full set. Row n = 4: sum of coefficients 1 C(4,0) + 4 C(4,1) + 6 C(4,2) + 4 C(4,3) + 1 C(4,4) = 16 set {A, B, C, D} — each element in or out — 2⁴ = 16 subsets ∅ {A},{B},... {A,B},... {A,B,C},... {A,B,C,D} 1 + 4 + 6 + 4 + 1 = 16 = 2⁴
The $16$ subsets of $\{A, B, C, D\}$ grouped by size. There are $\binom{4}{0} = 1$ with $0$ elements, $\binom{4}{1} = 4$ with $1$ element, and so on. The total is $1 + 4 + 6 + 4 + 1 = 16 = 2^4$.

Identity 2: Alternating sum of binomial coefficients

Alternating sum of binomial coefficients

For every positive integer n:

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

That is, \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \cdots = 0.

Algebraic proof. Set a = 1 and b = -1 in the binomial theorem:

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

The left side is 0^n = 0 (for n \geq 1). Done.

Combinatorial proof. Split the subsets of \{1, 2, \dots, n\} into two groups: those with an even number of elements and those with an odd number. The even-sized subsets contribute +\binom{n}{r} and the odd-sized subsets contribute -\binom{n}{r}. The identity says these two groups are equal in total count:

\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots

Both sides equal 2^{n-1}. You can prove this by constructing a bijection: pair each subset S with the subset S \triangle \{1\} (toggle element 1). This pairing maps even-sized subsets to odd-sized subsets and vice versa, so the two groups have the same count.

Alternating sum of row 4 coefficients equals zeroRow 4 coefficients with alternating signs: plus 1 minus 4 plus 6 minus 4 plus 1 equals 0. Below, even-sized subsets sum to 8 and odd-sized subsets sum to 8, showing the two groups are equal. Row n = 4: alternating sum +1 −4 +6 −4 +1 = 0 even-sized subsets C(4,0) + C(4,2) + C(4,4) = 1 + 6 + 1 = 8 odd-sized subsets C(4,1) + C(4,3) = 4 + 4 = 8 both = 2³ = 8, so alternating sum = 0
For $n = 4$: the even-indexed coefficients ($1 + 6 + 1 = 8$) equal the odd-indexed coefficients ($4 + 4 = 8$). Their difference is zero. Both groups total $2^{4-1} = 8$.

Identity 3: Sum of even-indexed and odd-indexed coefficients

From the two identities above, you can extract each half.

Add the sum identity and the alternating-sum identity:

2^n + 0 = 2\big[\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots\big]

So the sum of the even-indexed binomial coefficients is 2^{n-1}. Subtracting instead:

2^n - 0 = 2\big[\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots\big]

The sum of the odd-indexed binomial coefficients is also 2^{n-1}.

\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots = 2^{n-1}

Identity 4: Weighted sum \sum r \binom{n}{r} = n \cdot 2^{n-1}

This identity computes a different kind of sum — each coefficient is multiplied by its index r.

Proof. Start from (1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r. Differentiate both sides with respect to x:

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

Set x = 1:

n \cdot 2^{n-1} = \sum_{r=1}^{n} r\binom{n}{r} = \sum_{r=0}^{n} r\binom{n}{r}

(The r = 0 term is zero, so including it does not change the sum.)

Check for n = 4: 0 \cdot 1 + 1 \cdot 4 + 2 \cdot 6 + 3 \cdot 4 + 4 \cdot 1 = 0 + 4 + 12 + 12 + 4 = 32 = 4 \cdot 2^3. Confirmed.

Identity 5: Vandermonde's identity

Vandermonde's identity

For non-negative integers m, n, and p with p \leq m + n:

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

Combinatorial proof. You have two groups: one with m people and another with n people. Choose p people from the combined group. The left side counts this directly. The right side counts by splitting: choose k from the first group (\binom{m}{k} ways) and p - k from the second group (\binom{n}{p-k} ways), then sum over all possible k.

Check. \binom{4+3}{3} = \binom{7}{3} = 35. Sum: \binom{4}{0}\binom{3}{3} + \binom{4}{1}\binom{3}{2} + \binom{4}{2}\binom{3}{1} + \binom{4}{3}\binom{3}{0} = 1 \cdot 1 + 4 \cdot 3 + 6 \cdot 3 + 4 \cdot 1 = 1 + 12 + 18 + 4 = 35. Confirmed.

Interactive: verify identities with any row

Drag the red point to select a row n (from 1 to 8). The readouts show the row sum, the alternating sum, and the weighted sum for that row, verifying the three main identities.

Interactive verification of binomial coefficient identitiesA horizontal axis for n from 1 to 8. A draggable red point selects a row. Readouts show the sum of C(n,r) equalling 2 to the n, the alternating sum equalling 0, and the weighted sum equalling n times 2 to the n minus 1. n value 1 2 3 4 5 6 7 row sum 2ⁿ weighted sum n·2ⁿ⁻¹ ↔ drag the red point
Drag the point to select a row $n$. The upper curve plots $2^n$ (the row sum) and the lower curve plots $n \cdot 2^{n-1}$ (the weighted sum). The alternating sum is always $0$ regardless of $n$.

Summary of identities

Identity Formula Proof method
Row sum \displaystyle\sum_{r=0}^{n} \binom{n}{r} = 2^n Set a = b = 1
Alternating sum \displaystyle\sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0 Set a = 1, b = -1
Even/odd split Each half = 2^{n-1} Add/subtract the two above
Weighted sum \displaystyle\sum_{r=0}^{n} r\binom{n}{r} = n\cdot 2^{n-1} Differentiate, set x = 1
Vandermonde \displaystyle\binom{m+n}{p} = \sum_{k} \binom{m}{k}\binom{n}{p-k} Combinatorial counting

Two worked examples

Example 1: Find the value of $\binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \cdots + \binom{10}{10}$

Step 1. Recognise the pattern. This is the sum of all binomial coefficients for n = 10.

Why: the sum runs from r = 0 to r = 10 with no sign changes and no weights — it matches the row-sum identity exactly.

Step 2. Apply the identity \sum_{r=0}^{n} \binom{n}{r} = 2^n.

\binom{10}{0} + \binom{10}{1} + \cdots + \binom{10}{10} = 2^{10}

Why: this is the binomial theorem with a = b = 1: (1 + 1)^{10} = 2^{10}.

Step 3. Compute 2^{10}.

2^{10} = 1024

Why: 2^{10} = 1024 is a standard power of 2 worth memorising.

Step 4. Verify with a partial check. \binom{10}{0} + \binom{10}{1} + \binom{10}{2} = 1 + 10 + 45 = 56. The remaining terms must sum to 1024 - 56 = 968. Since \binom{10}{3} = 120 and the terms grow to \binom{10}{5} = 252, this is plausible.

Result. \displaystyle\sum_{r=0}^{10} \binom{10}{r} = 1024.

Row 10 of Pascal's triangle summing to 1024Eleven boxes showing the values C(10,0) through C(10,10): 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. A summation line below gives 1024 = 2 to the 10. The values form a symmetric hill shape. Row n = 10 1 10 45 120 210 252 210 120 45 10 1 symmetric: C(10, r) = C(10, 10−r) peak at r = 5: C(10, 5) = 252 sum = 1024 = 2¹⁰
The eleven coefficients of row $10$ form a symmetric hill, peaking at $\binom{10}{5} = 252$ in the centre. Their sum is $2^{10} = 1024$, the total number of subsets of a $10$-element set.

The hill shape — rising to a peak at r = n/2 and then falling symmetrically — is characteristic of binomial coefficients. The peak is always at r = n/2 (or the two values nearest n/2 when n is odd). This connects to the general and middle terms article's discussion of the greatest coefficient.

Example 2: Prove that $\binom{8}{0} - \binom{8}{1} + \binom{8}{2} - \cdots + \binom{8}{8} = 0$, and find $\binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8}$

Step 1. Apply the alternating-sum identity for n = 8.

\sum_{r=0}^{8} (-1)^r \binom{8}{r} = (1 - 1)^8 = 0

Why: setting a = 1, b = -1 in the binomial theorem gives (1 + (-1))^8 = 0^8 = 0 on the left, and the alternating sum on the right.

Step 2. Write the identity explicitly.

\binom{8}{0} - \binom{8}{1} + \binom{8}{2} - \binom{8}{3} + \binom{8}{4} - \binom{8}{5} + \binom{8}{6} - \binom{8}{7} + \binom{8}{8} = 0

That is: 1 - 8 + 28 - 56 + 70 - 56 + 28 - 8 + 1 = 0.

Why: the positive terms (1 + 28 + 70 + 28 + 1 = 128) exactly cancel the negative terms (8 + 56 + 56 + 8 = 128).

Step 3. Find the sum of even-indexed coefficients. Add the row sum (2^8 = 256) and the alternating sum (0):

256 + 0 = 2\big[\binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8}\big]
\binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} = \frac{256}{2} = 128

Why: when you add the row sum and the alternating sum, all the odd-indexed terms cancel (they appear with opposite signs), leaving twice the even-indexed sum.

Step 4. Verify: 1 + 28 + 70 + 28 + 1 = 128. And the odd-indexed sum: 8 + 56 + 56 + 8 = 128. Both halves equal 2^{8-1} = 2^7 = 128. Confirmed.

Result. The alternating sum is 0, and the sum of even-indexed coefficients is 128 = 2^7.

Even and odd indexed binomial coefficients for n equals 8 both sum to 128Nine coefficients of row 8 are split into two groups. Even-indexed (r = 0, 2, 4, 6, 8): 1 + 28 + 70 + 28 + 1 = 128. Odd-indexed (r = 1, 3, 5, 7): 8 + 56 + 56 + 8 = 128. Both equal 2 to the 7. Row n = 8: even vs odd split 1 8 28 56 70 56 28 8 1 r=0 r=1 r=2 r=3 r=4 r=5 r=6 r=7 r=8 even-indexed (red) 1 + 28 + 70 + 28 + 1 = 128 = 2⁷ odd-indexed 8 + 56 + 56 + 8 = 128 = 2⁷ alternating sum: 128 − 128 = 0
The nine coefficients of row $8$, coloured by parity of index. The even-indexed coefficients (red) sum to $128$ and the odd-indexed coefficients sum to $128$. Their difference is $0$ (the alternating-sum identity) and each half equals $2^7 = 128$.

The even-odd split is a recurring technique in binomial-coefficient problems. Any time a sum involves only the even-indexed or only the odd-indexed coefficients, the answer is 2^{n-1}. This fact, combined with the row sum 2^n, handles a large class of JEE-style questions.

Common confusions

Going deeper

If you can compute row sums, alternating sums, and weighted sums, and you understand Vandermonde's identity, you are ready for general and middle terms. The rest of this section collects a few more identities and a deeper perspective.

The absorption identity

The identity r\binom{n}{r} = n\binom{n-1}{r-1} is proved by writing out both sides:

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

This identity converts a factor of r in front of a binomial coefficient into a simpler expression. It is the algebraic engine behind the weighted sum \sum r\binom{n}{r} = n \cdot 2^{n-1}: replace r\binom{n}{r} with n\binom{n-1}{r-1}, pull the n out of the sum, and what remains is \sum \binom{n-1}{r-1} = 2^{n-1}.

The second-moment identity

Differentiating (1+x)^n twice and setting x = 1 gives:

\sum_{r=0}^{n} r(r-1)\binom{n}{r} = n(n-1) \cdot 2^{n-2}

Combined with the first-moment identity \sum r\binom{n}{r} = n \cdot 2^{n-1}, this lets you compute \sum r^2\binom{n}{r}:

\sum r^2 \binom{n}{r} = \sum r(r-1)\binom{n}{r} + \sum r\binom{n}{r} = n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1}

Generating functions perspective

Each identity in this article is a special case of a single idea: the binomial theorem defines a polynomial (1+x)^n = \sum \binom{n}{r} x^r, and evaluating or differentiating this polynomial at specific values of x extracts different sums of the coefficients. Setting x = 1 gives the total; setting x = -1 gives the alternating sum; differentiating and setting x = 1 gives the weighted sum. This is the generating-function approach — the polynomial encodes all the coefficient information, and algebraic operations on the polynomial decode it.

Where this leads next