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Binomial coefficients and Pascal's triangle

Identities, the binomial theorem, and combinatorial proofs.

C(n, k) counts k-subsets, is the coefficient in (a + b)ⁿ, and satisfies Pascal's rule C(n, k) = C(n−1, k−1) + C(n−1, k). Picture it: Pascal's triangle, each entry the sum of the two above. Think it: a combinatorial proof shows two formulas count the same set — no algebra needed.

作業例: expand (a + b)^6

Algebra · Expanding

Expand (a + b)^6

\left(a + b\right)^{6}

ステップバイステップ

  1. \left(a + b\right)^{6}

    Start from the expression.

  2. a^{6} + 6 a^{5} b + 15 a^{4} b^{2} + 20 a^{3} b^{3} + 15 a^{2} b^{4} + 6 a b^{5} + b^{6}

    Distribute (multiply out) the products.

答えを明らかにしろ
a^{6} + 6 a^{5} b + 15 a^{4} b^{2} + 20 a^{3} b^{3} + 15 a^{2} b^{4} + 6 a b^{5} + b^{6}

Symbols used here

\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
n!
factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.
\binom{n}{k}
binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).
\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
\prod_{k=1}^{n} a_k
product
Multiply a_k for k = 1 up to n.
\emptyset,\ |A|
empty set, cardinality
The set with no elements; the number of elements of A.

How to: Binomial coefficients and Pascal's triangle

  1. Start from the expression.
  2. Distribute (multiply out) the products.

Questions people ask

Permutation or combination?

Ask whether order matters. A lock code is a permutation (order matters); a hand of cards is a combination (it does not).

What is a graph in this sense?

Dots (vertices) joined by lines (edges) — not a plot. Road maps, social networks and molecules are graphs; questions like "is there a route" and "how few colours" are graph theory.

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ここに Combinatorics & Graph Theory

The counting principlesPigeonhole principle and inclusion–exclusionRecurrences and generating functionsGraphs: vertices, edges, degreesPaths, cycles, trees, Euler and HamiltonColouring and planar graphs