Combinations & Pascal's Triangle

Binomial coefficients, subsets, and the binomial theorem

Combinations and Pascal's Triangle

When order stops mattering, counting turns into choosing, and the numbers that come out build Pascal's triangle.

Choosing k from n

Pick n and k and look through every k-item selection. Compare C(n, k) with C(n, n-k): they always match.

Subsets of size 2 from 5 items — C(5,2) = 1012345All C(5,2) = 10 subsets:1.12345{1, 2}2.12345{1, 3}3.12345{1, 4}4.12345{1, 5}5.12345{2, 3}6.12345{2, 4}7.12345{2, 5}8.12345{3, 4}9.12345{3, 5}10.12345{4, 5}C(5,2) = 5! / (2! × 3!) = 10Symmetry: C(5,2) = C(5,3) = 10 ✓

Filled circles indicate selected items. Each row is one of the C(n,k) possible subsets.

Go deeper:Choosing k from n

While permutations care about order, combinations count unordered selections. "How many ways can you choose k items from n?" is answered by the binomial coefficient C(n, k) = n! / (k!(n-k)!). These numbers form Pascal's triangle, one of the most pattern-rich objects in all of mathematics.

The binomial coefficient C(n, k) counts the number of ways to choose k items from a set of n, where order does not matter. Notice the symmetry: C(n, k) = C(n, n-k), because choosing which k to include is equivalent to choosing which n-k to exclude.

Key insight: C(n, k) = C(n, n-k) is not just an algebraic identity -- it reflects a bijection: every k-element subset corresponds to a unique (n-k)-element complement.

Pascal's Triangle Explorer

Add rows and toggle the highlighting modes. Color by odd and even to see the Sierpinski triangle appear.

Each entry C(n,k) equals the sum of the two entries above it. Toggle highlight modes to reveal hidden patterns.

Go deeper:Pascal's Triangle Explorer

Pascal's triangle arranges binomial coefficients in rows: row n contains C(n, 0), C(n, 1), ..., C(n, n). Each entry equals the sum of the two entries above it. Toggle highlighting modes to discover hidden patterns: the Sierpinski triangle from odd/even coloring, modular arithmetic patterns, and diagonal sequences.

Key insight: Coloring Pascal's triangle by parity (odd = colored, even = dark) reveals the Sierpinski triangle fractal. This happens because C(n, k) mod 2 = 1 if and only if every bit of k is also a bit of n (Lucas' theorem).

The Binomial Theorem

Pick n and watch (a + b)ⁿ expand term by term, each coefficient lit up in Pascal's triangle.

(a + b)⁴ = Σ C(4,k) a... — expand below(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴a⁴+4a³b+6a²b²+4ab³+b⁴Pascal's Triangle — row 4 provides the coefficients111121133114641

Click any term to highlight it and its Pascal's triangle entry. Use auto-play to step through each term.

Go deeper:The Binomial Theorem

The binomial theorem states that (a + b)ⁿ = Σ C(n, k) · aⁿ⁻ᵏ · bᵏ. Each term corresponds to choosing which k of the n factors contribute b (and the rest contribute a). Watch the expansion unfold term by term, with each coefficient linked to its position in Pascal's triangle.

Key insight: The binomial theorem is not just a formula -- it is a counting statement. The coefficient C(n, k) counts the number of ways to pick k copies of b from n factors of (a + b).

Key Takeaways

  • C(n, k) counts unordered selections and satisfies the symmetry C(n, k) = C(n, n-k).
  • Pascal's triangle encodes binomial coefficients with additive recurrence C(n,k) = C(n-1,k-1) + C(n-1,k).
  • Hidden fractal and modular patterns emerge from simple coloring rules.
  • The binomial theorem connects polynomial algebra to combinatorial counting.

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