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Pascal's Triangle

Pascal's Triangle

The figure below, extended infinitely. A particular entry is found by adding the two numbers that are above and on either side of the element. Note: The numbers which make up Pascal's triangle are called binomial coefficients.

 

Pascal's Triangle

Note that the sum of any two adjacent elements in a row can be found between them on the next row. Each row begins and ends with 1.

  

etc.

 

See also

Binomial coefficients in Pascal's triangle, binomial theorem

Key Formula

(nk)=(n−1k−1)+(n−1k)\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}
Where:
  • nn = The row number (starting from row 0 at the top)
  • kk = The position within the row (starting from position 0 on the left)
  • (nk)\binom{n}{k} = The binomial coefficient 'n choose k', equal to the entry in row n, position k

Worked Example

Problem: Build the first 6 rows of Pascal's Triangle (rows 0 through 5), then find the entry in row 5, position 2.
Step 1: Row 0 has a single entry: just 1.
Row 0: 1\text{Row 0: } 1
Step 2: Row 1 also starts and ends with 1.
Row 1: 11\text{Row 1: } 1 \quad 1
Step 3: For row 2, the edges are 1 and the middle entry is 1 + 1 = 2.
Row 2: 121\text{Row 2: } 1 \quad 2 \quad 1
Step 4: Continue the pattern. Each interior entry is the sum of the two entries above it.
Row 3: 1331\text{Row 3: } 1 \quad 3 \quad 3 \quad 1
Step 5: Build rows 4 and 5 the same way.
Row 4: 14641Row 5: 15101051\text{Row 4: } 1 \quad 4 \quad 6 \quad 4 \quad 1 \qquad \text{Row 5: } 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1
Step 6: The entry at row 5, position 2 (counting from 0) is 10. Verify with the formula:
(52)=5!2! 3!=1202⋅6=10\binom{5}{2} = \frac{5!}{2!\,3!} = \frac{120}{2 \cdot 6} = 10
Answer: The entry in row 5, position 2 of Pascal's Triangle is 10.

Another Example

Problem: Use Pascal's Triangle to expand (a+b)4(a + b)^4.
Step 1: Row 4 of Pascal's Triangle gives the coefficients: 1, 4, 6, 4, 1.
Row 4: 14641\text{Row 4: } 1 \quad 4 \quad 6 \quad 4 \quad 1
Step 2: Each coefficient multiplies a term where the powers of a decrease and the powers of b increase.
(a+b)4=1 a4+4 a3b+6 a2b2+4 ab3+1 b4(a+b)^4 = 1\,a^4 + 4\,a^3b + 6\,a^2b^2 + 4\,ab^3 + 1\,b^4
Answer: (a+b)4=a4+4a3b+6a2b2+4ab3+b4(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

Frequently Asked Questions

What are the patterns hidden in Pascal's Triangle?
There are many. The first diagonal is all 1s, the second diagonal lists the counting numbers (1, 2, 3, …), and the third diagonal lists the triangular numbers (1, 3, 6, 10, …). The sum of each row equals a power of 2: row n sums to 2n2^n. If you shade only the odd entries, you get a fractal pattern resembling the Sierpiński triangle.
How do you find a specific number in Pascal's Triangle without building the whole thing?
Use the binomial coefficient formula directly: the entry in row nn, position kk is (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}. For example, row 10, position 3 is (103)=10!3! 7!=120\binom{10}{3} = \frac{10!}{3!\,7!} = 120.

Why It Matters

Pascal's Triangle provides the coefficients needed to expand any binomial expression (a+b)n(a + b)^n, making it essential for the Binomial Theorem. It also encodes the number of ways to choose kk items from nn items, so it is central to combinatorics and probability. Many surprising number patterns—Fibonacci numbers, powers of 2, triangular numbers—all emerge from this single structure.

Common Mistakes

Mistake: Starting the row and position numbering at 1 instead of 0.
Correction: Pascal's Triangle is conventionally numbered starting from row 0 and position 0. Row 0 contains the single entry 1. If you start at 1, your binomial coefficient formula (nk)\binom{n}{k} will point to the wrong entry.
Mistake: Forgetting that every row starts and ends with 1.
Correction: The outermost entries are always 1 because (n0)=1\binom{n}{0} = 1 and (nn)=1\binom{n}{n} = 1. Only the interior entries are found by adding two numbers from the row above.

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