Pascals triangle
Pascal's triangle is a key tool for combinatorics in Year 11 Specialist Mathematics in Queensland (QCAA): a triangular array in which every entry is the sum of the two entries above it.
You will learn to build each row by the sum rule, read its entries as combinations, and apply Pascal's rule and symmetry. Each row totals a power of two and counts the subsets of a set — groundwork for probability and later counting.
Theory
Pascal's triangle is a triangular array of numbers in which every entry is the sum of the two entries above it. In Year 11 Specialist Mathematics (QCAA, Queensland) its entries are the combinations \({}^{n}C_{r}\), so the triangle reveals Pascal's rule, the symmetry of \({}^{n}C_{r}\), and why each row adds to \(2^{n}\).
Pascal's triangle is a triangular array of whole numbers. Each row starts and ends with \(1\), and every entry in between is the sum of the two entries immediately above it (the sum rule). The rows are numbered from \(n=0\) at the top, and the positions within a row from \(r=0\) on the left.
The entry in row \(n\), position \(r\), is exactly the combination \({}^{n}C_{r}\) — the number of ways to choose \(r\) objects from \(n\) when order does not matter. So the triangle is just the combinations \({}^{n}C_{r}\) laid out row by row.
Because each entry equals the sum of the two above, the combinations obey Pascal's rule: \({}^{n}C_{r}={}^{n-1}C_{r-1}+{}^{n-1}C_{r}\). Each row is also symmetric, since \({}^{n}C_{r}={}^{n}C_{n-r}\), and the entries of row \(n\) add to \(2^{n}\).
That row total \(2^{n}\) counts every subset of an \(n\)-element set: the \({}^{n}C_{r}\) entry counts the subsets of size \(r\), and adding them across the row gives all \(2^{n}\) subsets.
The entry in row \(n\), position \(r\), is the combination
Each entry is the sum of the two above it, which is Pascal's rule:
Each row is symmetric, and the entries of row \(n\) add to a power of two:
How to use Pascal's triangle
- Write the row above, putting a \(1\) at each end.
- Add neighbouring pairs: each new inner entry is the sum of the two entries directly above it (the sum rule, i.e. Pascal's rule).
- Read off or check the entry you need — the entry in row \(n\), position \(r\), is \({}^{n}C_{r}\); use symmetry \({}^{n}C_{r}={}^{n}C_{n-r}\) to shorten the work.
- Add across a row when you want a total: the whole of row \(n\) sums to \(2^{n}\), the number of subsets of an \(n\)-set.
Keep a \(1\) at each end, then add each neighbouring pair of entries above:
| \(\text{ends}\) | \(=\) | \(1\) |
| \(\text{next}\) | \(=\) | \(1 + 4\) |
| \(=\) | \(5\) | |
| \(\text{middle}\) | \(=\) | \(4 + 6\) |
| \(=\) | \(10\) | |
| \(\text{middle}\) | \(=\) | \(6 + 4\) |
| \(=\) | \(10\) | |
| \(\text{next}\) | \(=\) | \(4 + 1\) |
| \(=\) | \(5\) | |
| \(\text{row } n=5\) | \(:\) | \(1,\ 5,\ 10,\ 10,\ 5,\ 1\) |
Row \(n=5\) is \(1,\ 5,\ 10,\ 10,\ 5,\ 1\).
Pascal's rule adds the two entries directly above:
| \({}^{7}C_{2}\) | \(=\) | \({}^{6}C_{1} + {}^{6}C_{2}\) |
| \({}^{6}C_{1}\) | \(=\) | \(6\) |
| \({}^{6}C_{2}\) | \(=\) | \(15\) |
| \({}^{7}C_{2}\) | \(=\) | \(6 + 15\) |
| \(=\) | \(21\) |
\({}^{7}C_{2} = 21\).
Add the entries one at a time, left to right:
| \(\text{sum}\) | \(=\) | \(1 + 6\) |
| \(=\) | \(7\) | |
| \(=\) | \(7 + 15\) | |
| \(=\) | \(22\) | |
| \(=\) | \(22 + 20\) | |
| \(=\) | \(42\) | |
| \(=\) | \(42 + 15\) | |
| \(=\) | \(57\) | |
| \(=\) | \(57 + 6\) | |
| \(=\) | \(63\) | |
| \(=\) | \(63 + 1\) | |
| \(=\) | \(64\) | |
| \(2^{6}\) | \(=\) | \(64\) |
The entries add to \(64 = 2^{6}\).
Each element is independently either in or out of a subset — \(2\) choices each — so multiply across the \(5\) elements:
| \(\text{choices per element}\) | \(=\) | \(2\) |
| \(\text{total subsets}\) | \(=\) | \(2^{5}\) |
| \(=\) | \(2 \times 2 \times 2 \times 2 \times 2\) | |
| \(=\) | \(32\) |
Check against row \(n=5\) of Pascal's triangle (subsets grouped by size):
| \(\text{by size}\) | \(=\) | \(1 + 5 + 10 + 10 + 5 + 1\) |
| \(=\) | \(32\) |
There are \(32\) subsets, matching \(2^{5}=32\).
Common pitfalls
Frequently asked questions
How do you build Pascal's triangle?
Start each row with a \(1\) at both ends, then make every inner entry the sum of the two entries directly above it. Row \(n=4\) is \(1,4,6,4,1\), so row \(n=5\) is \(1,5,10,10,5,1\).
What is Pascal's rule?
Pascal's rule is \({}^{n}C_{r}={}^{n-1}C_{r-1}+{}^{n-1}C_{r}\). It says each combination equals the sum of the two combinations directly above it in the triangle.
Why does each entry equal \({}^{n}C_{r}\)?
The entry in row \(n\), position \(r\), counts the ways to choose \(r\) objects from \(n\), which is \({}^{n}C_{r}\). The sum rule matches the way choices split into "include the last object or not".
Why do the entries in a row add up to \(2^{n}\)?
Row \(n\) lists the number of subsets of each size, and every element of an \(n\)-set is independently in or out of a subset, giving \(2^{n}\) subsets in total.
Why is Pascal's triangle symmetric?
Choosing \(r\) objects to include is the same as choosing the \(n-r\) objects to leave out, so \({}^{n}C_{r}={}^{n}C_{n-r}\) and each row reads the same left to right as right to left.