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  1. Home
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  4. /Pascal's Triangle Generator

Pascal's Triangle Generator

Last updated: March 28, 2026

Calculator

Results

C(n, 0)

1

C(n, 1)

6

C(n, 2)

15

C(n, 3)

20

C(n, 4)

15

C(n, 5)

6

C(n, 6)

1

Row Sum = 2ⁿ

64

Central Element (if n even)

20

Alternating Sum

1

Results

C(n, 0)

1

C(n, 1)

6

C(n, 2)

15

C(n, 3)

20

C(n, 4)

15

C(n, 5)

6

C(n, 6)

1

Row Sum = 2ⁿ

64

Central Element (if n even)

20

Alternating Sum

1

The Pascal's Triangle Generator computes the first seven elements of any row in Pascal's triangle — the infinite triangular array of binomial coefficients. Row $$n$$ contains the values $$\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$$, which are the coefficients in the binomial expansion $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$.

Pascal's triangle is named after Blaise Pascal, who published a comprehensive treatise Traité du triangle arithmétique in 1665. However, the triangle was known centuries earlier: Chinese mathematician Jia Xian described it around 1050 CE, Persian mathematician Omar Khayyam discussed it in the 11th century, and it appears in Yang Hui's 1261 text. In China, it is still called Yang Hui's triangle; in Iran, it is Khayyam's triangle.

Each entry is the binomial coefficient:

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

The defining property of Pascal's triangle is the addition rule: each entry equals the sum of the two entries directly above it — $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$. This recursive identity, combined with the boundary conditions $$\binom{n}{0} = \binom{n}{n} = 1$$, generates the entire triangle.

Pascal's triangle encodes a stunning wealth of number-theoretic and combinatorial information. The row sums are powers of 2: $$\sum_{k=0}^{n}\binom{n}{k} = 2^n$$. The alternating row sums vanish for $$n \geq 1$$: $$\sum(-1)^k \binom{n}{k} = 0$$. The diagonal sums produce the Fibonacci sequence. The central binomial coefficients $$\binom{2n}{n}$$ are connected to Catalan numbers and Wallis's product for $$\pi$$.

In probability theory, row $$n$$ gives the coefficients of the binomial distribution $$B(n, 1/2)$$. In algebra, the binomial theorem extends to negative and fractional exponents via the generalized binomial series. In combinatorics, $$\binom{n}{k}$$ counts the number of ways to choose $$k$$ items from $$n$$ — the foundation of enumeration theory. Kummer's theorem relates the highest power of a prime dividing $$\binom{n}{k}$$ to the number of carries in base-$$p$$ addition, and Lucas's theorem gives $$\binom{n}{k} \bmod p$$ via the base-$$p$$ digits of $$n$$ and $$k$$.

Visual Analysis

How It Works

The calculator computes binomial coefficients using the multiplicative formula for efficiency and numerical stability.

Step 1: Apply the product formula. Rather than computing factorials (which overflow quickly), use:

$$\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} = \prod_{i=1}^{k} \frac{n - k + i}{i}$$

For example: $$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$.

Step 2: Generate C(n, 0) through C(n, 6).

$$\binom{n}{0} = 1, \quad \binom{n}{1} = n, \quad \binom{n}{2} = \frac{n(n-1)}{2}$$

$$\binom{n}{3} = \frac{n(n-1)(n-2)}{6}, \quad \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24}$$

$$\binom{n}{5} = \frac{n(n-1)(n-2)(n-3)(n-4)}{120}, \quad \binom{n}{6} = \frac{n(n-1)\cdots(n-5)}{720}$$

Step 3: Row sum. $$\sum_{k=0}^{n}\binom{n}{k} = 2^n$$ (set $$a = b = 1$$ in the binomial theorem).

Step 4: Central binomial coefficient. For even $$n = 2m$$, compute $$\binom{2m}{m}$$ using Stirling's approximation.

Understanding Your Results

C(n, k) values are the binomial coefficients — row $$n$$ of Pascal's triangle. Each value $$\binom{n}{k}$$ counts the number of ways to choose $$k$$ elements from a set of $$n$$.

Row Sum = 2ⁿ confirms the identity $$\sum \binom{n}{k} = 2^n$$. This is the total number of subsets of an $$n$$-element set.

Central Element is $$\binom{n}{n/2}$$ for even $$n$$ — the largest entry in the row and the peak of the binomial distribution. It grows as $$2^n / \sqrt{\pi n/2}$$.

Alternating Sum is $$\sum (-1)^k \binom{n}{k}$$, which equals 0 for all $$n \geq 1$$ (set $$a = 1, b = -1$$ in the binomial theorem). For $$n = 0$$, it equals 1.

Worked Examples

Row 6 of Pascal's Triangle

Inputs

n6

Results

c01
c16
c215
c320
c415
c56
c61
rowSum64
centralBinom20
altSum0

Row 6: 1, 6, 15, 20, 15, 6, 1. The row is symmetric (C(6,k) = C(6,6−k)). Row sum = 64 = 2⁶. The central element C(6,3) = 20 is the largest entry.

Row 10: Larger Coefficients

Inputs

n10

Results

c01
c110
c245
c3120
c4210
c5252
c6210
rowSum1024
centralBinom252
altSum0

Row 10: 1, 10, 45, 120, 210, 252, 210, ... The central binomial coefficient C(10,5) = 252 is the maximum. Row sum = 1024 = 2¹⁰.

Frequently Asked Questions

The symmetry $$\binom{n}{k} = \binom{n}{n-k}$$ follows directly from the formula: $$\frac{n!}{k!(n-k)!} = \frac{n!}{(n-k)!k!}$$. Combinatorially, choosing $$k$$ items to include is the same as choosing $$n-k$$ items to exclude. This symmetry means each row reads the same forwards and backwards.

The diagonals contain remarkable sequences: the first diagonal is all 1s, the second is the natural numbers, the third is the triangular numbers, the fourth is tetrahedral numbers. Shallow diagonal sums produce the Fibonacci sequence. Coloring entries by divisibility by a prime $$p$$ produces Sierpinski-triangle-like fractals. Hockey-stick identities, Vandermonde's identity, and many other combinatorial results can be read directly from the structure.

Row $$n$$ gives the number of ways to get exactly $$k$$ heads in $$n$$ fair coin flips: $$P(k \text{ heads}) = \binom{n}{k}/2^n$$. This is the binomial distribution $$B(n, 1/2)$$. For general success probability $$p$$, $$P(k) = \binom{n}{k}p^k(1-p)^{n-k}$$. As $$n \to \infty$$, this approaches the normal distribution — the central limit theorem.

Lucas's theorem states that for a prime $$p$$, $$\binom{n}{k} \equiv \prod_{i} \binom{n_i}{k_i} \pmod{p}$$ where $$n_i$$ and $$k_i$$ are the base-$$p$$ digits of $$n$$ and $$k$$. This gives a fast way to compute $$\binom{n}{k} \bmod p$$ and explains fractal patterns in Pascal's triangle modulo primes.

The calculator displays $$\binom{n}{0}$$ through $$\binom{n}{6}$$ — the first 7 elements. For rows with $$n > 12$$, these elements represent the left half of the row. Since Pascal's triangle is symmetric ($$\binom{n}{k} = \binom{n}{n-k}$$), the remaining elements can be read in reverse order from the displayed values when $$n \leq 12$$.

The binomial series $$$(1+x)^\alpha = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^k$$$ generalizes the binomial theorem to non-integer exponents $$\alpha$$, where $$\binom{\alpha}{k} = \frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}$$. This gives Taylor expansions like $$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \cdots$$ and is fundamental to analysis and approximation theory.

Sources & Methodology

Pascal, B., Traité du triangle arithmétique, 1665. Graham, R., Knuth, D. & Patashnik, O., Concrete Mathematics, 2nd ed., Addison-Wesley, 1994. Koshy, T., Triangular Arrays with Applications, Oxford, 2011. OEIS, Sequence A007318 (Pascal's triangle).
R

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