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Definition of Binomial theorem, properties of binomial coefficients

binomial theorem

The Binomial theorem, also known as binomial expansion, explains the expansion of powers. It only applies to binomials. So, let’s start…

If we know that

(a + b)^2 = 1 * a + 1 * b

(a + b)^2 = 1 * a^2 + 2 * a * b + 1* b^2

(a + b)^3 = 1 * a^3 + 3 * a^2 * b + 3 * a * b^2 + 1 * b^3

(a + b)^4 = 1 * a^4 + 4 * a^3 * b + 6 * a^2 * b^2 + 4 * a * b^3 + 1 * b^4

And so on.

Notice that all addends come in the form b_i * a^{n - i} * b^i. Numbers b_i are called binomial coefficients. They are easily calculated and noted using factorials. Factorial function is a function that multiplies first n natural numbers. For natural number n with n! we denote the multiplication of first n natural numbers.

n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1

1! = 1
2! = 2 * 1
3! = 3 * 2 * 1
4! = 4 * 3 * 2 * 1
5! = 5 * 4 * 3 * 2 * 1
6! = 6 * 5 * 4 * 3 * 2 * 1
Also, notice that the following recursion is also true:

n! = n * (n - 1)! also, 0! = 1 by the agreement.

 

Example 1. Calculate \frac{49! - 48!}{48!}

First, when dealing with factorials never jump ahead and calculate everything because you can always somehow use the recursion. Now since we know that 49! = 49 * 48! we have a common factor of al terms.

\frac{49 * 48! - 48!}{48!} = \frac{48! (49 - 1)}{48!} = 48

 

Example 2. Solve the following equation \frac{n!}{(n - 3)!} = \frac{8 (n - 1)!}{(n - 2)!}

\frac{n * (n - 1) * (n - 2) * (n - 3)!}{(n - 3)!} = \frac{8 (n - 1) (n - 2)!}{(n - 2)!}

n * (n - 1) * (n - 2) = 8 * (n - 1)

n * (n - 2) = 8

n^2 - 2n - 8 = 0

n = 4

Note: Second solution of this quadratic equation is removed because n has to be natural number.

 

Binomial coefficients

If n is a natural number and k is a natural number or 0 and k \le n binomial coefficient is denoted with symbol {n}\choose{k} and defined as:

{n}\choose{k} =\frac{n!}{k! (n - k)!}

The binomial coefficients have some specific properties.

1. {n}\choose{0} =  1

Proof. When replacing k with 0 in the definition of binomial coefficient we get:

{n}\choose{0} = \frac{n!}{0! (n - 0!)} = \frac{n!}{1 * n!} = \frac{n!}{n!} = 1

 

2. The symmetry property:

{n}\choose{k} =  {n}\choose{n - k}

Proof.

{n}\choose{k} =  {n}\choose{k! (n - k)!}

 {n}\choose{n - k} = \frac{n!}{(n - k)!(n - (n - k))!} = \frac{n!}{(n - k)! k!}

 

3.  {n}\choose{k} + {n}\choose{k - 1} = \frac{n!}{k!(n - k)!} + \frac{n!}{(k - 1)!(n - k + 1)!} = \frac{n!}{(k - 1)!(n - k)!} * (\frac{1}{k} * \frac{1}{(k - 1)!(n - k + 1)!} ) = \frac{n!(n + 1)}{(k - 1)!(n - k)!(n - k + 1)k} =  {n + 1}\choose{k}

 

Example 3. Calculate {8}\choose{6} using these properties.

{8}\choose{6} = {8}\choose{8 - 6} = {8}\choose{2} = \frac{8 * 7}{2 * 1} = 28

 

Example 4. Calculate {14}\choose{12}{14}\choose{11} using these properties.

{14}\choose{12} + {14}\choose{11} = {15}\choose{12} = {13}\choose{3} = 455

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