Quick answer
Row n, position k (from 0) equals C(n, k).
Formula
- C(n,k) = C(n−1,k−1) + C(n−1,k)
- Row n: C(n,0)…C(n,n)
Introduction
Draw the first few rows by hand. The symmetry jumps off the page: each row reads the same forward and backward because C(n, k) = C(n, n − k).
When you need algebra connections, follow binomial theorem and coefficients to see how row n matches the expansion of (a + b)^n.
If factorials feel heavy, compare triangle entries with the closed-form formula for the same n and k until both routes agree.
Triangle basics
Row 0 is a single 1. Row 1 is 1, 1. Row 2 is 1, 2, 1. In general, row n has n + 1 entries counting k = 0 through n.
Construct row n from row n − 1 by adding neighbors. End entries are always 1 because C(n, 0) and C(n, n) equal 1.
Path interpretation: walking from the top with down-left and down-right steps reaches position k after n moves with exactly k right turns, linking counting to lattice paths.
Recursive property
- C(n,k) = C(n−1,k−1) + C(n−1,k) for 0 < k < n
The recursion is the same identity used in inductive proofs of the binomial theorem. Computers use dynamic programming on this idea when n is moderate and you need many coefficients at once.
Step-by-step guide
- Locate row n. Start counting rows at 0 unless your instructor labels differently; note the convention once.
- Find column k. Leftmost entry is k = 0.
- Cross-check with formula. For exam safety, verify one entry per row using the product method.
Row 6 in two ways
Triangle construction gives 1, 6, 15, 20, 15, 6, 1.
Formula check: C(6,3) = 20 matches the middle entry.
Sum of row n equals 2^n because (1 + 1)^n counts all subsets; coefficients sum to the same total.
