Generate Pascal’s triangle to any depth, read off binomial coefficients, and reveal hidden patterns — with every row explained. Free and mobile-friendly.
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Pascal’s triangle is a triangular array of numbers where each entry is the sum of the two directly above it. It begins 1; 1 1; 1 2 1; 1 3 3 1 — and from this humble rule springs one of the richest structures in mathematics, encoding combinations, binomial expansions, probabilities, and even fractal art. Our free Pascal’s Triangle Calculator generates the triangle to any depth up to 15 rows, highlights odd or even entries to expose hidden patterns, and explains what each row means.
Named for Blaise Pascal (1623–1662), who systematized its properties in his Treatise on the Arithmetical Triangle, the pattern was actually known centuries earlier to Chinese mathematician Yang Hui and Persian scholar Omar Khayyam. Its endurance across cultures testifies to something deep: the triangle is not a curiosity but a crossroads where arithmetic, algebra, combinatorics, and probability all meet.
This page unpacks the triangle completely: the one-line construction rule, why each row gives binomial coefficients (the numbers in (a + b)ⁿ), how the rows sum to powers of 2, where the Fibonacci numbers hide, how coloring odd entries draws the Sierpiński fractal, and how coin-flip probabilities fall straight out of row 4. Worked examples, practice problems, and eight FAQs round out the tour.
The construction rule could not be simpler: start with 1 at the top, and every subsequent entry is the sum of the two numbers above it (treating missing neighbors as 0). Row 0 is just [1]. Row 1 is [1, 1]. Row 2: the middle entry 1 + 1 = 2, giving [1, 2, 1]. Row 3: [1, 3, 3, 1]. Each row has one more entry than the last, and the edges are always 1 — because 1 + 0 = 1 all the way down the sides.
The deep meaning: the k-th entry of row n is the binomial coefficient C(n, k) = n! ÷ (k!(n−k)!), the number of ways to choose k items from n. Row 4 reads 1 4 6 4 1 — there is 1 way to choose nothing from 4, 4 ways to choose one, 6 ways to choose two, and so on. This is why the triangle answers probability questions: the entries of row n count the outcomes of n yes/no events.
The triangle also encodes binomial expansions: row n gives the coefficients of (a + b)ⁿ. Row 2 (1 2 1) gives (a + b)² = a² + 2ab + b²; row 3 (1 3 3 1) gives (a + b)³ = a³ + 3a²b + 3ab² + b³. Before the triangle, expanding (a + b)ⁿ meant tedious multiplication; after it, you read the coefficients off the triangle and write the answer.
Entry: C(n, k) = n! ÷ (k! × (n−k)!) • Row sum: Σ row n = 2ⁿ • (a + b)ⁿ from row n
Rows are numbered from n = 0 at the top; positions k run from 0 at the left edge. So “row 4, position 2” is the 6 in 1 4 6 4 1.
Generate 5 rows (n = 0…4). Row 4 reads 1 4 6 4 1. As combinations: C(4,0) = 1, C(4,1) = 4, C(4,2) = 6, C(4,3) = 4, C(4,4) = 1. Check the middle: 4! ÷ (2! × 2!) = 24 ÷ 4 = 6 ✓. As a binomial expansion: (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴. Row sum: 1 + 4 + 6 + 4 + 1 = 16 = 2⁴ ✓.
Flip a fair coin 4 times. How many outcomes have exactly 2 heads? The answer sits in row 4: position 2 holds 6 — sequences like HHTT, HTHT, THHT, and three more. With 2⁴ = 16 total outcomes, P(exactly 2 heads) = 6 ÷ 16 = 37.5%. The whole probability distribution for 4 flips is row 4 divided by 16: 1/16, 4/16, 6/16, 4/16, 1/16.
Row 5 is 1 5 10 10 5 1. Each entry of row 6 sums adjacent pairs: 1; 1+5 = 6; 5+10 = 15; 10+10 = 20; 10+5 = 15; 5+1 = 6; 1 — giving 1 6 15 20 15 6 1. Verify with factorials: C(6,3) = 720 ÷ (6 × 6) = 720 ÷ 36 = 20 ✓. The additive rule and the factorial formula always agree; the triangle is just the fast lane.
A club with 8 members must choose a 3-person committee. How many possible committees? Read row 8, position 3 of the triangle: generate 9 rows and find 56. By factorial: C(8,3) = 8! ÷ (3! × 5!) = 40320 ÷ (6 × 120) = 40320 ÷ 720 = 56 ✓. Event planners, lottery designers, and poker analysts all reach for exactly this number.
Quality control uses the same row differently. A factory samples 8 chips and wants the probability that at most 1 is defective (defect rate 10% per chip). The binomial probabilities come from row 8 weighted by powers: P(0) = C(8,0)(0.9)⁸ ≈ 0.4305, P(1) = C(8,1)(0.1)(0.9)⁷ = 8 × 0.1 × 0.4783 ≈ 0.3826, so P(≤1) ≈ 81.3%. The triangle supplies the combinatorial skeleton; the probabilities dress it.
Sum the shallow diagonals of the triangle — lines sloping gently down-right, each starting one row lower — and you get 1, 1, 2, 3, 5, 8, 13, …: the Fibonacci sequence. This is not numerology; both structures obey the same additive recurrence, so the Fibonacci numbers are literally woven into the triangle’s fabric. Try it on the generator: diagonal sums of rows 0–6 give 1, 1, 2, 3, 5, 8.
Color every odd entry (use the widget’s highlight) and a famous fractal emerges: the Sierpiński triangle, an infinitely self-similar pattern of nested triangles. The reason is modular arithmetic — an entry is odd exactly when the corresponding binomial coefficient is odd, and Lucas’s theorem ties that to the binary digits of n and k. Finite rows show the first iterations; infinite rows would show the full fractal.
More patterns: the second diagonal (1, 3, 6, 10, 15, …) holds the triangular numbers; the third diagonal holds tetrahedral numbers. Row sums give powers of 2 (1, 2, 4, 8, 16, …). Alternating sums of any row equal zero. The hockey-stick identity says a diagonal sum equals the entry below the diagonal’s end. The triangle is a whole combinatorics course in one picture.
Compute 11⁴ without a calculator: read row 4 — 1 4 6 4 1 — and concatenate: 14641. This works because 11ⁿ = (10 + 1)ⁿ expands with binomial coefficients as decimal digits, as long as no coefficient reaches 10 (it breaks at row 5, where carries appear: 161051, not 15101051).
Add down diagonal 1 + 3 + 6 + 10 (starting row 2): the sum is 20, which equals the entry just below and left of the last term — row 6, position 3 is 20 ✓. The “hockey stick” (shaft down the diagonal, blade to the entry) always lands exactly.
Write row 7 of Pascal’s triangle using the addition rule. (Answer: 1 7 21 35 35 21 7 1.)
Expand (x + 2)³ using row 3. (Row 3 is 1 3 3 1 → x³ + 3·x²·2 + 3·x·4 + 8 = x³ + 6x² + 12x + 8.)
How many ways can you choose 2 toppings from 6? (Row 6, position 2: C(6,2) = 15.)
Before Pascal’s systematization, expanding (a + b)ⁿ for large n meant multiplying polynomials by hand — error-prone and slow. The triangle turned it into a lookup: row n is the coefficient list. Newton later extended the idea to non-integer exponents with the generalized binomial series, the engine behind countless approximations in physics and engineering. A 17th-century triangle still powers 21st-century computation.
Modern probability leans on it constantly. Every binomial distribution — coin flips, defect counts, survey responses — is a triangle row weighted by pᵏ(1−p)ⁿ⁻ᵏ. When your phone’s autocorrect weighs word probabilities or a poll reports a margin of error, binomial coefficients from this triangle are doing quiet work underneath.
The Pascal’s Triangle Calculator generates the triangle to a chosen depth (up to 15 rows) via the rule that each entry sums the two above it, with optional odd/even highlighting. Row n, position k equals the binomial coefficient C(n,k) = n!÷(k!(n−k)!); row n also gives the coefficients of (a+b)ⁿ, and every row sums to 2ⁿ. Related patterns: Fibonacci numbers on shallow diagonals, the Sierpiński fractal in odd entries, triangular numbers on the second diagonal.
A triangular array where each number is the sum of the two directly above it, starting with 1 at the top. Row n encodes the binomial coefficients C(n,k), the building blocks of combinations, binomial expansions, and binomial probabilities.
Write 1 at the top. For each new row, keep 1s on both edges and fill each interior entry with the sum of the two entries above it. Row 2 becomes 1 2 1; row 3 becomes 1 3 3 1; continue as far as needed.
Row n gives the coefficients of (a + b)ⁿ and the counts C(n,k) for choosing k items from n. Row 4 (1 4 6 4 1) expands (a+b)⁴ and counts the outcomes of 4 yes/no trials.
Row n sums to 2ⁿ because each entry splits into two entries below, doubling the total each row — equivalently, it counts all subsets of an n-element set, and there are 2ⁿ of those.
Summing the shallow diagonals (each starting one row lower, sloping down-right) yields 1, 1, 2, 3, 5, 8, 13, … — the Fibonacci sequence. Both structures share the same additive recurrence.
Coloring only the odd entries reveals the Sierpiński triangle fractal. An entry is odd exactly when its binomial coefficient is odd (Lucas’s theorem), producing infinite self-similar nesting.
Row n counts the outcomes of n binary trials: for 4 coin flips, row 4 (1 4 6 4 1) over 16 total outcomes gives P(2 heads) = 6/16 = 37.5%. It is the combinatorial core of the binomial distribution.
Summing down any diagonal of the triangle equals the entry just below and to the side of the diagonal’s end — e.g., 1 + 3 + 6 + 10 = 20. The pattern looks like a hockey stick, and it always works.
Cite this page as: Pascal’s Triangle Calculator. Sigma Calculator. Retrieved October 11, 2026, from https://sigmacalculator.net/math-calculators/pascals-triangle-calculator/
Last updated: October 11, 2026