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Combinatorics and Probability
The Binomial Theorem
Algebra Textbooks Boundless Algebra Combinatorics and Probability The Binomial Theorem
Algebra Textbooks Boundless Algebra Combinatorics and Probability
Algebra Textbooks Boundless Algebra
Algebra Textbooks
Algebra
Concept Version 10
Created by Boundless

Binomial Expansion and Factorial Notation

The binomial theorem describes the algebraic expansion of powers of a binomial.

Learning Objective

  • Use factorial notation to find the coefficients of a binomial expansion


Key Points

    • According to the theorem, it is possible to expand the power (x+y)n(x + y)^n(x+y)​n​​ into a sum involving terms of the form axbycax^by^cax​b​​y​c​​, where the exponents bbb and ccc are nonnegative integers with b+c=nb+c=nb+c=n, and the coefficient aaa of each term is a specific positive integer depending on nnn and bbb.
    • The factorial of a non-negative integer nnn, denoted by n!n!n!, is the product of all positive integers less than or equal to nnn.
    • Binomial coefficients can be written as (nk)\displaystyle{\begin{pmatrix} n \\ k \end{pmatrix}}(​n​k​​)or nCk_{n}{C}_{k} ​n​​C​k​​ and are defined in terms of the factorial function n!n!n!.

Terms

  • factorial

    The result of multiplying a given number of consecutive integers from 111 to the given number. In equations, it is symbolized by an exclamation mark (!!!). For example, 5!=1⋅2⋅3⋅4⋅5=1205! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 1205!=1⋅2⋅3⋅4⋅5=120.

  • binomial coefficient

    A coefficient of any of the terms in the expansion of the binomial power (x+y)n(x+y)^n(x+y)​n​​.


Full Text

Recall that the binomial theorem is an algebraic method of expanding a binomial that is raised to a certain power, such as (4x+y)7(4x+y)^7(4x+y)​7​​. The theorem is given by the formula: 

(x+y)n=∑k=0n(nk)xn−kyk=∑k=0n(nk)xkyn−k\displaystyle { (x+y) }^{ n }=\sum _{ k=0 }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } { x }^{ n-k }{ y }^{ k }=\sum _{ k=0 }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } { x }^{ k }{ y }^{ n-k }(x+y)​n​​=​k=0​∑​n​​(​n​k​​)x​n−k​​y​k​​=​k=0​∑​n​​(​n​k​​)x​k​​y​n−k​​

The coefficients that appear in the binomial expansion are called binomial coefficients. These are usually written (nk)\displaystyle{\begin{pmatrix} n \\ k \end{pmatrix}}(​n​k​​)or nCk_{n}{C}_{k}​n​​C​k​​, and pronounced "nnn choose kkk".

The coefficient of a term $x^{n−k}y^k$ in a binomial expansion can be calculated using the combination formula. Recall that the combination formula represents the number of ways to choose kkk objects from among nnn, where order does not matter. The formula consists of factorials:

(nk)=n!k!(n−k)!\displaystyle \begin{pmatrix} n \\ k \end{pmatrix} = \frac { n! }{ k!(n-k)! }(​n​k​​)=​k!(n−k)!​​n!​​

Note that although this formula involves a fraction, the binomial coefficient (nk)\begin{pmatrix} n \\ k \end{pmatrix}(​n​k​​) is actually an integer.

In calculating coefficients, recall that the factorial of a non-negative integer nnn, denoted by n!n!n!, is the product of all positive integers less than or equal to nnn. For example, $5!=5×4×3×2×1=120$. The value of 0!0!0! is 111, according to the convention for an empty product. 

Finally, you may recall that the factorial n!n!n! also represents the number of possible permutations of nnn, or the number of ways in which nnn objects can be arranged or selected.

Example: Use the binomial formula to find the expansion of (x+y)4(x+y)^4(x+y)​4​​

Start by substituting n=4n=4n=4 into the binomial formula:

(x+y)4=∑k=04(4k)x4−kyk\displaystyle (x+y)^4=\sum _{ k=0 }^{ 4 }{ \begin{pmatrix} 4 \\ k \end{pmatrix} } { x }^{ 4-k }{ y }^{ k }(x+y)​4​​=​k=0​∑​4​​(​4​k​​)x​4−k​​y​k​​

In order to solve this, we will need to expand the summation for all values of kkk.

(x+y)4=(40)x4−0y0+(41)x4−1y1+(42)x4−2y2+(43)x4−3y3+(44)x4−4y4=(40)x4+(41)x3y+(42)x2y2+(43)x1y3+(44)y4\displaystyle \begin{aligned} (x+y)^4&={ \begin{pmatrix} 4 \\ 0 \end{pmatrix} } { x }^{ 4-0}{y}^{0} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 4-1}{ y }^{1} + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 4-2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }^{ 4-3 }{ y }^{ 3 } + { \begin{pmatrix} 4 \\ 4 \end{pmatrix} } { x }^{ 4-4 }{ y }^{ 4 } \\ &={ \begin{pmatrix} 4 \\ 0 \end{pmatrix} } { x }^{ 4} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 3}{ y } + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }^{ 1 }{ y }^{ 3 } + { \begin{pmatrix} 4 \\ 4 \end{pmatrix} } { y }^{ 4 } \end{aligned}​(x+y)​4​​​​​​=(​4​0​​)x​4−0​​y​0​​+(​4​1​​)x​4−1​​y​1​​+(​4​2​​)x​4−2​​y​2​​+(​4​3​​)x​4−3​​y​3​​+(​4​4​​)x​4−4​​y​4​​​=(​4​0​​)x​4​​+(​4​1​​)x​3​​y+(​4​2​​)x​2​​y​2​​+(​4​3​​)x​1​​y​3​​+(​4​4​​)y​4​​​​

Recall that (40){ \begin{pmatrix} 4 \\ 0 \end{pmatrix} }(​4​0​​) and (44){ \begin{pmatrix} 4 \\ 4 \end{pmatrix} }(​4​4​​)are both equivalent to 1, as there is only one way to choose either 000 or 444 objects from among 444. Therefore, we have:

 =x4+(41)x3y+(42)x2y2+(43)xy3+y4\displaystyle = { x }^{ 4} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 3}{ y } + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }{ y }^{ 3 } + { y }^{ 4 } =x​4​​+(​4​1​​)x​3​​y+(​4​2​​)x​2​​y​2​​+(​4​3​​)xy​3​​+y​4​​

Now we must evaluate each of the remaining combinations: 

(41)=4!1!(4−1)!=4!1!3!=4\displaystyle \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \frac { 4! }{ 1!(4-1)! } = \frac { 4! }{ 1!3! } = 4(​4​1​​)=​1!(4−1)!​​4!​​=​1!3!​​4!​​=4

(42)=4!2!(4−2)!=4!2!2!=6\displaystyle \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \frac { 4! }{ 2!(4-2)! } = \frac { 4! }{ 2!2! } = 6(​4​2​​)=​2!(4−2)!​​4!​​=​2!2!​​4!​​=6

(43)=4!3!(4−3)!=4!3!1!=4\displaystyle \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \frac { 4! }{ 3!(4-3)! } = \frac { 4! }{ 3!1! } = 4(​4​3​​)=​3!(4−3)!​​4!​​=​3!1!​​4!​​=4

Substituting these integers into the expansion, we have:

(x+y)4=x4+4x3y+6x2y2+4xy3+y4\displaystyle (x+y)^4 = { x }^{ 4} + 4 { x }^{ 3}{ y } + 6 { x }^{ 2 }{ y }^{ 2 } + 4 { x }{ y }^{ 3 } + { y }^{ 4 } (x+y)​4​​=x​4​​+4x​3​​y+6x​2​​y​2​​+4xy​3​​+y​4​​

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