Pascal's Triangle Generator

About Pascal's Triangle Generator

How Is Pascal's Triangle Constructed?

Start with a 1 at the top. Each row begins and ends with 1. Every other entry is the sum of the two entries directly above it. Here are the first 8 rows:

Row (n)	Entries	Row Sum
0	1	1
1	1, 1	2
2	1, 2, 1	4
3	1, 3, 3, 1	8
4	1, 4, 6, 4, 1	16
5	1, 5, 10, 10, 5, 1	32
6	1, 6, 15, 20, 15, 6, 1	64
7	1, 7, 21, 35, 35, 21, 7, 1	128

Notice the row sums: 1, 2, 4, 8, 16, 32, 64, 128 - each is a power of 2. The sum of row n is always 2^n. This follows from the binomial theorem with a = b = 1: (1+1)^n = 2^n.

Binomial Coefficients - the C(n, r) Connection

The entry at row n, position r (counting from 0) equals the binomial coefficient C(n, r), calculated as n! / (r! x (n-r)!). Click any number in the triangle and this tool shows the full formula and calculation.

Worked example: What is the entry at row 6, position 2?

C(6, 2) = 6! / (2! x 4!) = 720 / (2 x 24) = 720 / 48 = 15

Check the table above - row 6 reads 1, 6, 15, 20, 15, 6, 1, and position 2 is indeed 15.

Property	Formula	Meaning
Symmetry	C(n, r) = C(n, n-r)	Choosing r items is the same as leaving out n-r items
Edges are 1	C(n, 0) = C(n, n) = 1	One way to choose nothing or everything
Pascal's rule	C(n, r) = C(n-1, r-1) + C(n-1, r)	Each entry = sum of two above it
Row sum	Sum of row n = 2^n	Total subsets of an n-element set
Alternating sum	C(n,0) - C(n,1) + C(n,2) - ... = 0	For n > 0, even and odd subsets balance

For computing individual combinations and permutations, the Permutations and Combinations Calculator handles larger values directly.

How Is Pascal's Triangle Used for Binomial Expansion?

The binomial theorem says (a + b)^n = sum of C(n, r) x a^(n-r) x b^r for r = 0 to n. The coefficients come straight from row n of Pascal's triangle.

Example: Expand (x + y)^4. Row 4 gives coefficients 1, 4, 6, 4, 1:

(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

Expression	Row	Expansion
(a+b)^0	0	1
(a+b)^1	1	a + b
(a+b)^2	2	a^2 + 2ab + b^2
(a+b)^3	3	a^3 + 3a^2b + 3ab^2 + b^3
(a+b)^4	4	a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4
(a+b)^5	5	a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

This is why Pascal's triangle is taught alongside algebra - it gives you the coefficients for any power of a binomial without multiplying everything out by hand.

Patterns Hidden in the Triangle

Pascal's triangle contains a surprising number of patterns. Use the highlight buttons in this tool to see them visually.

Pattern	Where to Find It	What You See
Natural numbers	Second diagonal (positions 1)	1, 2, 3, 4, 5, 6, 7...
Triangular numbers	Third diagonal (positions 2)	1, 3, 6, 10, 15, 21...
Tetrahedral numbers	Fourth diagonal (positions 3)	1, 4, 10, 20, 35...
Fibonacci numbers	Shallow diagonals (sum entries along)	1, 1, 2, 3, 5, 8, 13...
Powers of 2	Row sums	1, 2, 4, 8, 16, 32...
Powers of 11	Rows (concatenate single digits)	1, 11, 121, 1331, 14641...
Sierpinski triangle	Colour odd entries	Fractal triangular pattern
Catalan numbers	C(2n,n) / (n+1)	1, 1, 2, 5, 14, 42, 132...

The Fibonacci diagonal is one of the most elegant: sum the entries along each shallow diagonal going from the left edge upward to the right, and you get the Fibonacci sequence. For example, in the diagonal starting at row 5 position 0: C(5,0) + C(4,1) + C(3,2) = 1 + 4 + 3 = 8 = F(6). Explore the Fibonacci Generator to study the sequence directly.

The Sierpinski Triangle Fractal

Colour every odd entry one shade and every even entry another. What emerges is the Sierpinski triangle - a self-similar fractal that appears at every scale as you add more rows. This happens because C(n, r) is odd exactly when every bit of r is also a bit of n (in binary). The pattern becomes more dramatic as you generate more rows. At 16 rows you can see the first clear nesting; at 32 rows the fractal structure is unmistakable.

Pascal's Triangle in Probability

If you flip a fair coin n times, the number of ways to get exactly r heads is C(n, r). The probability of exactly r heads in n flips is C(n, r) / 2^n. Row n of Pascal's triangle gives the numerators, and 2^n (the row sum) gives the denominator.

Example: Flip a coin 5 times. Row 5 is 1, 5, 10, 10, 5, 1. Total outcomes = 32.

Heads	Ways C(5,r)	Probability
0	1	1/32 = 3.1%
1	5	5/32 = 15.6%
2	10	10/32 = 31.3%
3	10	10/32 = 31.3%
4	5	5/32 = 15.6%
5	1	1/32 = 3.1%

This is the binomial distribution in its simplest form. For applied probability problems, the Probability Calculator handles compound events and conditional probability.

Hockey Stick Identity

Pick any entry on the left edge (a 1), move diagonally down-right through several entries, then turn the corner one step straight down. The entry at the corner equals the sum of all the diagonal entries above it. This forms a hockey stick shape in the triangle.

C(r,r) + C(r+1,r) + C(r+2,r) + ... + C(n,r) = C(n+1, r+1)

Example: C(2,2) + C(3,2) + C(4,2) + C(5,2) = 1 + 3 + 6 + 10 = 20 = C(6,3).

Computing Large Binomial Coefficients

C(n, r)	Value	Context
C(10, 3)	120	Ways to pick 3 from 10
C(20, 10)	184,756	Middle of row 20
C(30, 15)	155,117,520	Middle of row 30
C(52, 5)	2,598,960	Possible 5-card poker hands
C(49, 6)	13,983,816	UK Lotto combinations
C(100, 50)	~1.009 x 10^29	Largest entry in row 100

The middle entry of each row is always the largest, and it grows roughly as 4^n / sqrt(pi x n) for large n. This rapid growth is why the numbers in Pascal's triangle get very large very quickly.

A Short History of the Triangle

Pascal's triangle is named after French mathematician Blaise Pascal, who published his Traité du triangle arithmétique in 1665, but the pattern was known centuries earlier around the world. Indian mathematician Pingala described it around 200 BCE in his work on Sanskrit prosody. Persian mathematician Al-Karaji wrote about it in the 10th century, and Omar Khayyam studied its properties around 1100 CE - it is still called Khayyam's triangle in Iran. In China, Jia Xian described it in the 11th century and Zhu Shijie recorded the first eight rows in his 1303 work Jade Mirror of the Four Unknowns, where it is known as Yang Hui's triangle. Pascal's contribution was connecting the triangle to probability theory through his correspondence with Pierre de Fermat on games of chance - a foundation the field of probability was built on.

Mathematician	Year	Region
Pingala	c. 200 BCE	India
Al-Karaji	c. 1000 CE	Persia
Omar Khayyam	c. 1100 CE	Persia
Jia Xian / Yang Hui	c. 1050-1261 CE	China
Zhu Shijie	1303 CE	China
Blaise Pascal	1665 CE	France

How Does This Tool Compute Large Entries?

Computing C(n, r) naively as n! / (r! x (n-r)!) overflows JavaScript numbers quickly, since 21! already exceeds 2^53. This tool uses the multiplicative formula C(n, r) = product of (n-i)/(i+1) for i from 0 to r-1, with the smaller of r and n-r chosen as the loop bound thanks to the symmetry rule. That keeps intermediate values small enough to survive within JavaScript's double-precision floats up to about row 30, where the largest entry C(30, 15) = 155,117,520 is still well inside the safe integer range of 2^53 - 1 (roughly 9.007 x 10^15). Beyond that, exact integer arithmetic would require BigInt - which is why the tool caps at 30 rows. For combinatorics work at larger n, a symbolic math system like Wolfram Alpha or a language with arbitrary-precision integers (Python, Haskell) is the right choice.

The triangle also builds entire rows recursively via Pascal's rule, C(n, r) = C(n-1, r-1) + C(n-1, r). That approach is useful when you need every entry in a row, because each entry is computed from adjacent entries already known - no factorials or divisions required. Most textbook implementations use this approach for teaching, while the multiplicative formula is preferred when you need a single binomial coefficient quickly. For related counting problems, the Factorial Calculator and Probability Calculator pair well with this tool.

Common Mistakes When Using Pascal's Triangle

The most frequent error is counting rows and positions starting from 1 instead of 0. The top row with a single 1 is row 0, and the first entry of every row is position 0. Mislabelling leads to off-by-one errors in every binomial coefficient lookup. Another common mistake is forgetting that C(n, r) and C(n, n-r) are equal - students often compute both instead of using symmetry to halve the work. Finally, the "powers of 11" pattern breaks at row 5 because the entries become two digits (1, 5, 10, 10, 5, 1) and carries have to be performed to recover 11^5 = 161,051. The pattern still holds if you interpret the row as a polynomial in 10 with no carries rather than as a concatenated integer.

Sources

• Encyclopaedia Britannica - Pascal's triangle
• Wolfram MathWorld - Pascal's Triangle
• Wolfram MathWorld - Binomial Coefficient
• MacTutor History of Mathematics - Blaise Pascal
• OEIS A007318 - Pascal's triangle read by rows
• Wolfram MathWorld - Sierpinski Sieve

All rendering runs in your browser. No data is sent to any server.