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Math · College algebra · Concept

Why binomial coefficients appear

An expansion counts choices from repeated factors.

Choose one term from every factor

In (a+b)^n there are n identical factors. Each product term chooses either a or b from each factor. To obtain a^(n−k)b^k, choose exactly k factors to supply b.

(a+b)n=∑k=0n(nk)an−kbk

Count choices without ordering them

The chosen positions form a subset, not an ordered list. There are n!/(k!(n−k)!) such subsets. This explains both the coefficient and the symmetry between k and n−k.

(nk)=n!k!(n−k)!

Pascal’s recurrence splits the possibilities

Separate subsets according to whether they include the last position. Those that include it choose k−1 from n−1; those that omit it choose k from n−1. Their counts add. This finite theorem assumes a nonnegative integer n.

(nk)=(n−1k−1)+(n−1k)

Common mistakes

  • Do not drop the mixed terms.
  • In (a−b)^n, substitute −b and preserve its alternating powers.

Work through an example

Use coefficients, powers and signs together.

Expand (x−2)³ →
Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.