6 Binomial Theorem

The Binomial Theorem is useful in developing theory around the Binomial and Hypergeometric Distributions. Two proofs of the Theorem are provided here; one using the traditional approach, and one using a more general approach. Other useful theorems are provided at the end of this chapter.

6.1 Traditional Proof

6.1.1 Lemma: Pascal’s rule

Let \(n\) and \(x\) be non-negative integers such that \(x\leq n\).

Then \({n-1\choose x} + {n-1\choose x-1} = {n\choose x}\).


\[\begin{aligned} {n-1\choose x} + {n-1\choose x-1} &= \frac{(n-1)!}{x!(n-1-x)!} + \frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}\\ &= \frac{(n-1)!}{x!(n-x-1)!} + \frac{(n-1)!}{(x-1)!(n-1-x+1)!}\\ &= \frac{(n-1)!}{x!(n-x-1)!} + \frac{(n-1)!}{(x-1)!(n-x)!}\\ &= \frac{(n-1)!}{x(x-1)!(n-x-1)!} + \frac{(n-1)!}{(x-1)!(n-x)(n-x-1)!}\\ &= \frac{x(n-1)!}{x(x-1)!(n-x)(n-x-1)!} +\frac{(n-x)(n-1)!}{x(x-1)!(n-x)(n-x-1)!}\\ &= \frac{x(n-1)!+(n-x)(n-1)!}{x(x-1)!(n-x)(n-x-1)!} \\ &= \frac{(x+n-x)(x-1)!}{x(x-1)!(n-x)(n-x-1)!}\\ &= \frac{n(n-1)!}{x(x-1)!(n-x)(n-x-1)!} \\ &= \frac{n!}{x!(n-x)!} \\ &= {n\choose x} \end{aligned}\]

6.1.2 The Binomial Theorem

Let \(a\) and \(b\) be constants and let \(n\) be any positive integer. Then

\[(a+b)^n = \sum\limits_{x=0}^{n} {n\choose x} a^{n-x} b^x\]


This proof is completed by mathematical induction.

Base Step: \(n=1\)

\[\begin{aligned} (a+b)^1 &= \sum\limits_{x=0}^{1} {1\choose x} a^{1-x} b^x \\ &= {1\choose 0} a^{1-0} b^0 + {1\choose 1} a^{1-1} b^1 \\ &= 1\cdot a\cdot 1 + 1\cdot 1\cdot b \\ &= a+b \end{aligned}\]

Inductive Step: Assume that the Theorem holds for \(n\), and show it is true for \(n+1\).

\[\begin{aligned} (a+b)^{n+1} &= (a+b)(a+b)^n \\ &= a(a+b)^n + b(a+b)^n \\ &= a(a^n + \sum\limits_{x=1}^{n-1}{n\choose x}a^{n-x}b^x + b^n) + b(a^n + \sum\limits_{x=1}^{n-1}{n\choose x}a^{n-x}b^x+b^n) \\ &= (a^{n+1}+a\sum\limits_{x=1}^{n-1}{n\choose x}a^{n-x}ab^x) + (a^nb+\sum\limits_{x=1}^{n-1}{n\choose x}a^{n-x}b^x+b^{n+1}) \\ &= (a^{n+1}+\sum\limits_{x=1}^{n-1}{n\choose x}a^{n-x+1}ab^x) + (a^nb+\sum\limits_{x=1}^{n-1}{n\choose x}a^{n-x}b^{x+1}+b^{n+1}) \\ ^{[1]} &= (a^{n+1}+\sum\limits_{x=1}^{n}a^{n-x+1}b^x) + (\sum\limits_{x=0}^{n-1}{n\choose x}a^{n-x}b^{x+1}+b^{n+1}) \\ ^{[2]} &= (a^{n+1}+\sum\limits_{x=1}^{n}{n\choose x}a^{n-x+1}b^x) + \sum\limits_{x-1}^{n-1}{n\choose x-1}a^{n-x+1}b^{x+1-1}+b^{n+1}) \\ ^{[3]} &= a^{n+1} + \sum\limits_{x+1}^{n}{n+1\choose x}a^{n-x+1}b^x + b^{n+1} \\ &=a^{n+1}+\sum\limits_{x=1}^{n}{n+1\choose x}a^{(n+1)-x}b^x+b^{n+1} \\ ^{[4]} &= \sum\limits_{x=0}^{n+1}{n+1\choose x}a^{(n+1)-x}b^x \end{aligned}\]

This completes both the inductive step and the proof.

  1. \(ab^n={n\choose n}a^{n-n+1}b^n\) which is the term for \(x=n\) in the first summation.
    \(a^nb={n\choose 0}a^{n-0}b^1\) which is the term for \(x=0\) in the second summation.
  2. \(\sum\limits_{x=0}^{n-1}{n\choose x}a^{n-x}b^{x+1} \\ \ \ \ \ = \sum\limits_{x=1}^{n}{n\choose x-1}a^{n-(x-1)}b^{(x-1)+1} \\ \ \ \ \ = \sum\limits_{x=1}^{n}{n\choose x-1}a^{n-x+1}b^x\)
  3. This step is made using Pascal’s Rule with \(n=n-1\).
  4. \(a^{n+1}={n+1\choose 0}a^{(n+1)-0}b^0\) which is the term for \(x=0\) in the summation.
    \(\ \ b^{n+1}={n+1\choose n+1}a^{(n+1)-(n+1)}b^{n+1}\) which is the term for \(x=n+1\) in the summation

6.2 General Approach

6.2.1 A Binomial Expansion Theorem

This theorem and its corrolary are provided by Brunette (Brunette 2003–2007a).

For any positive integer \(n\), let \(B_n = (x_1+y_1) (x_2+y_2) \cdots (x_n+y_n)\). In the expansion \(B_n\), before combining possible like terms, the following are true:

  1. There will be \(2^n\) terms.
  2. Each of these terms will be a product of \(n\) factors.
  3. In each such product there will be one factor from each binomial (in \(B_n\)).
  4. Every such product of \(n\) factors, one from each binomial, is represented in the expansion.


Proof is done by induction.

For the case \(n=1\), the result is clear.

Now assume that the theorem is true for a particular \(n\) and consider \(B_{n+1}\).

\[ B_{n+1} = B_n(x_{n+1} + y_{n+1}) = B_nx_{n+1} + B_ny_{n+1} \]

By the inductive assumption, \(B_n = T_1 + T_2 + \cdots + T_{2^n}\) where each \(T_i\) is a product of \(n\) factors, one factor from each binomial. It follows that every term in the expansion of \(B_n+1\) is either of the type \(T_ix_{n+1}\) or \(T_iy_{n+1}\), for some \(1\leq i \leq 2^n\). But each term of either of the above types is clearly a product of \(n+1\) factors with one factor coming from each binomial. thus, if (ii) and (iii) are true for \(B_n\), then they are true for \(B_n+1\).

Next, by the inductive assumption, the expansion of \(B_n\) is a sum of \(2^n+2^n\) terms, i.e., \(2^{n+1}\) terms. This completes the inductive step for (i).

Lastly, it remains for us to consider a product of the type \(p_1 p_2 \cdots p_n p_{n+1}\) where, for each \(1\leq i\leq n+1\), \(p_i = x_i\) or \(p_i = y_i\). By the inductive hypothesis, \(p_1 p_2 \cdots p_n\) is a term in the expansion of \(B_n\). If \(p_{n+1} = x_{n+1}\), then \(p_1 p_2 \cdots p_n p_{n+1}\) is a term in the expansion of \(B_nx_{n+1}\), and so of \(B_{n+1}\). Likewise, if \(p_{n+1}=y_{n+1}\), then \(p_1 p_2 \cdots p_n p_{n+1}\) is a term in the expansion of \(B_n y_{n+1}\), and so of \(B_{n+1}\). This completes the inductive step and the proof.

6.2.2 Corollary: Binomial Theorem

Let \(x\) and \(y\) be constants and let \(n\) be any positive integer.

Then \(\displaystyle (x+y)^n = \sum\limits_{i=0}^{n} {n\choose i} x^{n-i} y^i\\\)


Since each term in the expansion will have \(n\) terms, each term must follow the form \(x^{n-i} y^i\) for \(0 \leq i \leq n\), and in all, there are \(2^n\) such terms. For any given value of \(i\), the number of terms of the form \(x^{n-i}y^i\) is clearly the number of ways one can choose the \(i\) factors of \(y\) from the \(n\) available binomials, i.e., \({n\choose i}\), which gives

\[(x+y)^n = \sum\limits_{i=0}^{n}{n\choose i} x^{n-i} y^i\]

6.3 Other Theorems

6.3.1 Theorem

\[{N_1\choose 0}{N_2\choose n} + {N_1\choose 2}{N_2\choose n-1} + \cdots + {N_1\choose n-1}{N_2\choose 1} + {N_1\choose n}{N_2\choose 0} = {N_1+N_2\choose n}\]

where \(0 \leq n \leq N_1 + N_2\).


Using the Binomial Theorem we establish

\[ (1+a)^{N-1} (1+a)^{N_2} = (1+a)^{N_1+N_2} \\ \Rightarrow [{N_1\choose 0}a^0+\cdots+{N_1\choose N_1}a^{N_1}]\cdot [{N_2\choose 0}a^0+\cdots+{N_2\choose N_2}a^{N_2}] \\ \ \ \ \ ={N_1+N_2\choose 0}+{N_1+N_2\choose 1}a+\cdots +{N_1+N_2\choose N_1+N_2}a^{N_1+N_2} \]

Expanding the left side of the equation gives

\[ {N_1\choose 0}{N_2\choose 0} + {N_1\choose 0}{N_2\choose 1}a + \cdots + {N_1\choose 0}{N_2\choose N_2}a^{N_2} + {N_1\choose 1}{N_2\choose 0}a \\ \ \ \ \ + \cdots + {N_1\choose 1}{N_2\choose N_2}a^{N_2+1} + \cdots + {N_1\choose N_1}{N_2\choose 0}a^{N_1} + {N_1\choose N_1}{N_2\choose 1}a^{N_1+1} \\ \ \ \ \ + \cdots + {N_1\choose N_1}{N_2\choose N_2}a^{N_1+N_2} \\ = {N_1\choose 0}{N_2\choose 0}+{N_1\choose 0}{N_2\choose 1}a + {N_1\choose 1}{N_2\choose 0}a \\ \ \ \ \ + {N_1\choose 0}{N_2\choose 2}a^2+{N_1\choose 1}{N_2\choose 1}a^2 + {N_1\choose 2}{N_2\choose 0}a^2 \\ \ \ \ \ + \cdots + {N_1\choose N_1}{N_2\choose N_2}a^{N_1+N_2} \]

Notice that for any \(n\) where \(0 \leq n \leq N_1 + N_2\), the coefficient for \(a^n\), found by combining like terms, is \({N_1\choose 0}{N_2\choose n} + {N_1\choose 1}{N_2\choose n-1} + \cdots+{N_1\choose n-1}{N_2\choose 1} + {N_1\choose 0}{N_2\choose n}\) and, by the equivalence of the first equation in the proof, is equal to the coefficient \({N_1 + N_2\choose n}\).

6.3.2 Theorem

\[\frac{\sum\limits_{i=1}^{n}{N_1\choose i}{N_2\choose n-i}}{{N_1+N_2\choose n}} = 1\]

for \(0 \leq n \leq N_1 + N_2\).\


Theorem 6.3.1 establishes the equality

\[ {N_1\choose 0}{N_2\choose n}+{N_1\choose 2}{N_2\choose n-1} + \cdots + {N_1\choose n-1}{N_2\choose 1}+{N_1\choose n}{N_2\choose 0} = {N_1+N_2\choose n} \\ \Rightarrow\sum\limits_{i=1}^{n}{N_1\choose i}{N_2\choose n-i} = {N_1+N_2\choose n} \\ \Rightarrow\frac{\sum\limits_{i=1}^{n} {N_1\choose i}{N_2\choose n-i}} {{N_1+N_2\choose n}} = 1 \]


Brunette, John. 2003–2007a. “A Binomial Expansion Theorem with the Binomial Theorem as a Corrolary.” Lecture Notes.