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Binomial Theorem

Binomial Theorem

Consider the power $(a + x)^n$

  • $(a + x)^2 = a^2 + 2xa + x^2 = aa + ax + xa + xx$
  • $(a + x)^3 = aaa + aax + axa + axx + xaa + xax + xxa + xxx$

These formulas contain all permutations with repetitions of the symbols $x$ and $a$

  • $(a + x)^n$ is the same thing: it contains all possible permutations of $a$ and $x$ of length $n$

Let us find how many terms there are that contain $k$ symbols $x$ and $n - k$ symbols $a$

  • $P(k, n - k) = C_n^k = \frac{n!}{k! \cdot (n - k)!}$ (the number of permutations with repetitions for two groups equals the number of combinations of size $k$ out of $n$)
  • i.e. we take the term $x^k \cdot a^{n - k}$ with the coefficient $C_n^k$ and get
  • $(a + x)^n = C_n^0 a^n + C_n^1 a^{n-1} x + … + C_n^k a^{n-k} x^k + … + C_n^n x^n$

This formula is known as the Binomial Theorem

General case

For $(x_1 + … + x_m)^n$ the coefficient of $x_1^{k_1} \cdot x_2^{k_2} \cdot … \cdot x_m^{k_m}$ is $P(k_1, k_2, …, k_m)$.

Proof of the Properties of Combinations

Call a function of the form $(1 + x)^n$ a generating function

$(1 + x)^n = C_n^0 + C_n^1 x + … + C_n^k x^k + … + C_n^n x^n$.

With this formula it is easy to prove the properties of combinations, in particular property 3 and property 6

Property 3: $\sum_{k = 0}^n C_n^k = 2^n$

  • Let $x$ in the generating function be 1. Then
  • $2^n = C_n^0 + C_n^1 + … + C_n^k + … + C_n^n$.

Property 6: $C_n^0 - C_n^1 + … + (-1)^n C_n^k = 0$

  • Let $x = -1$. Then
  • $0 = C_n^0 - C_n^1 + … + (-1)^n C_n^k$.

See also

Sources

  • Vilenkin N.Ya., Combinatorics. Moscow: Nauka, 1969. (in Russian)