Binomial Coefficients Calculator | iCalculator™ (2024)

The Binomial Coefficients Calculator will calculate:

1. The coefficients of any binomial when both terms and the power of the binomial are given

Binomial Coefficients Calculator Parameters: The power of the binomial is a natural (counting) number.

Please note that the formula for each calculation along with detailed calculations is shown further below this page. As you enter the specific factors of each binomial coefficients calculation, the Binomial Coefficients Calculator will automatically calculate the results and update the formula elements with each element of the binomial coefficients calculation. You can then email or print this binomial coefficients calculation as required for later use.

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Theoretical Description

A binomial is a mathematical expression raised to a certain power n where the expression itself contains two terms combined with each other through the operation of addition according to the scheme below.

Binomial = (Term 1 + Term 2)ⁿ

The term "binomial" means a "polynomial containing two terms". The power in which a polynomial is raised is called its degree or order. Thus, a binomial raised to the second power is called a second-order (degree) binomial, when raised to the third power it is called a third-order (degree) binomial, etc.

Finding the coefficients of a binomial means finding all numbers preceding the variables in each of the terms when the binomial is written in the decomposed form. In general, we express the terms using the letters x and y (or a and b). For example, in binomials raised to the second power (quadratic binomials), these coefficients are 1, 2 and 1 respectively, because

(a + b)² = a² + 2ab + b²

Likewise, the coefficients in the binomials raised to the third power (cubic binomials) are 1, 3, 3 and 1 respectively, because

(a + b)³ = a³ + 3a² b + 3ab² + b³

The coefficients of the second and third-order binomials can be found using the expanding brackets method. However, higher order binomials become harder to calculate through the expanding brackets method, so we must use other methods to calculate them. Therefore, we must use a more comprehensive method, which allows the calculation of binomial coefficients of any degree. This method is called the binomial coefficients theorem.

This theorem, first discovered by Sir Isaac Newton, says that the coefficients preceding the variables in binomials raised to a given power are as follows:

(a + b)ⁿ = (n0) ∙ aⁿ ∙ b⁰ + (n1) ∙ a^{n - 1} ∙ b¹ + (n2) ∙ a^{n - 2} ∙ b² + ⋯ + (nk) ∙ a^{n - k} ∙ b^k + ⋯ + (nn - 1) ∙ a¹ ∙ b^{n - 1} + (nn) ∙ a⁰ ∙ bⁿ

The general term of this binomial expression therefore is

k^th term = (nk) a^{n - k} b^k

where

(nk) = n!/k!(n - k)!

The symbol "!" is for "factorial". It means multiplying a number n by all the other numbers from n to 1, i.e. n! = n × (n - 1) × (n - 2) × × [n - (n - 1)]. For example,

5! = 5 · 4 · 3 · 2 · 1
8! = 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1

etc.

Hence, the algebraic form of expansion of the binomial expression (a + b)ⁿ is

(a + b)ⁿ = ⁿ∑_{k = 0}(nk) a^{n - k} b^k

where the symbol ⁿ∑_{k = 0} is an abbreviation that means "the sum of all terms from k = 0 to k = n".

From the above formula and from the definition of factorial, it is clear that the first and the last coefficients are both 1, because

(n0) = C(n,0) = n!/0!(n-0)! = n!/1 ∙ n! = 1

and

(nn) = C(n,n) = n!/n!(n - n)! = n!/n! ∙ 0! = n!/n! ∙ 1 = 1

Look at the two examples below that show how to find the binomial coefficients for n = 6 and for n = 7.

For n = 6, (k varies from 0 to 6) we have

(a + b)⁶=(60) ∙ a⁶ ∙ b⁰ + (61) ∙ a^{6 - 1} ∙ b¹ + (62) ∙ a^{6 - 2} ∙ b² + (63) ∙ a^{6 - 3} ∙ b³ + (64) ∙ a^{6 - 4} ∙ b⁴ + (65) ∙ a^{6 - 5} ∙ b⁵ + (66) ∙ a^{6 - 6} ∙ b⁶
= 6!/0!(6 - 0)! ∙ a⁶ ∙ b⁰ + 6!/1!(6 - 1)! ∙ a⁵ ∙ b¹ + 6!/2!(6 - 2)! ∙ a⁴ ∙ b² + 6!/3!(6 - 3)! ∙ a³ ∙ b³ + 6!/4!(6 - 4)! ∙ a² ∙ b⁴ + 6!/5!(6 - 5)! ∙ a¹ ∙ b⁵ + 6!/6!(6 - 6)! ∙ a⁰ ∙ b⁶
= 6!/0! ∙ 6! ∙ a⁶ ∙ b⁰ + 6!/1! ∙ 5! ∙ a⁵ ∙ b¹ + 6!/2! ∙ 4! ∙ a⁴ ∙ b² + 6!/3! ∙ 3! ∙ a³ ∙ b³ + 6!/4! ∙ 2! ∙ a² ∙ b⁴ + 6!/5! ∙ 1! ∙ a¹ ∙ b⁵ + 6!/6! ∙ 0! ∙ a⁰ ∙ b⁶
= 6!/1 ∙ 6! ∙ a⁶ ∙ 1 + 6 ∙ 5!/1 ∙ 5! ∙ a⁵ ∙ b + 6 ∙ 5 ∙ 4!/2 ∙ 1 ∙ 4! ∙ a⁴ ∙ b² + 6 ∙ 5 ∙ 4 ∙ 3!/3 ∙ 2 ∙ 1 ∙ 3! ∙ a³ ∙ b³ + 6 ∙ 5 ∙ 4!/4! ∙ 2 ∙ 1 ∙ a² ∙ b⁴ + 6 ∙ 5!/5! ∙ 1 ∙ a¹ ∙ b⁵ + 6!/6! ∙ 1 ∙ 1 ∙ b⁶
= a⁶ + 6a⁵ b + 15a⁴ b² + 20a³ b³ + 15a² b⁴ + 6ab⁵ + b⁶

For n = 7 (k varies from 0 to 7), we have

(a + b)⁷=(70) ∙ a⁷ ∙ b⁰ + (71) ∙ a^{7 - 1} ∙ b¹ + (72) ∙ a^{7 - 2} ∙ b² + (73) ∙ a^{7 - 3} ∙ b³ + (74) ∙ a^{7 - 4} ∙ b⁴ + (75) ∙ a^{7 - 5} ∙ b⁵ + (76) ∙ a^{7 - 6} ∙ b⁶ + (77) ∙ a^{7 - 7} ∙ b⁷
= 7!/0!(7 - 0)! ∙ a⁷ ∙ b⁰ + 7!/1!(7 - 1)! ∙ a⁶ ∙ b¹ + 7!/2!(7 - 2)! ∙ a⁵ ∙ b² + 7!/3!(7 - 3)! ∙ a⁴ ∙ b³ + 7!/4!(7 - 4)! ∙ a³ ∙ b⁴ + 7!/5!(7 - 5)! ∙ a² ∙ b⁵ + 7!/6!(7 - 6)! ∙ a¹ ∙ b⁶ + 7!/7!(7 - 7)! ∙ a⁰ ∙ b⁷
= 7!/0! ∙ 7! ∙ a⁷ ∙ b⁰ + 7!/1! ∙ 6! ∙ a⁶ ∙ b¹ + 7!/2! ∙ 5! ∙ a⁵ ∙ b² + 7!/3! ∙ 4! ∙ a⁴ ∙ b³ + 7!/4! ∙ 3! ∙ a³ ∙ b⁴ + 7!/5! ∙ 2! ∙ a² ∙ b⁵ + 7!/6! ∙ 1! ∙ a¹ ∙ b⁶ + 7!/7! ∙ 0! ∙ a⁰ ∙ b⁷
= 7!/1 ∙ 7! ∙ a⁷ ∙ 1 + 7 ∙ 6!/1! ∙ 6! ∙ a⁶ ∙ b + 7 ∙ 6 ∙ 5!/2 ∙ 1 ∙ 5! ∙ a⁵ ∙ b² + 7 ∙ 6 ∙ 5 ∙ 4!/3 ∙ 2 ∙ 1 ∙ 4! ∙ a⁴ ∙ b³ + 7 ∙ 6 ∙ 5 ∙ 4!/4! ∙ 3 ∙ 2 ∙ 1 ∙ a³ ∙ b⁴ + 7 ∙ 6 ∙ 5!/5! ∙ 2 ∙ 1 ∙ a² ∙ b⁵ + 7 ∙ 6!/6! ∙ 1 ∙ a ∙ b⁶ + 7!/7! ∙ 1 ∙ 1 ∙ b⁷
= a⁷ + 7a⁶ b + 21a⁵ b² + 35a⁴ b³ + 35a³ b⁴ + 21a² b⁵ + 7ab⁶ + b⁷

Instructions and information for using the Binomial Coefficients Calculator

All you have to do is to insert the two terms a and b of the binomial as well as the index (exponent) n that shows the power of the binomial. The terms a and b are not meant to be just single letters; they can also be monomials. For example, you may insert 2x for a and 3y for b. The calculator will eventually list all the binomial coefficients when the original binomial is written in the disassembled form.

For example, if you insert 2x for a, 4y for b and 3 for n, the calculator gives the following coefficients:

Output 1 = 8; Output 2 = 48; Output 3 = 96; Output 4 = 64

because

(a + b)³ = (30) ∙ a³ ∙ b⁰ + (31) ∙ a^{3 - 1} ∙ b¹ + (32) ∙ a^{3 - 2} ∙ b² + (33) ∙ a^{3 - 3} ∙ b³
= 3!/0!(3-0)! ∙ a³ ∙ b⁰ + 3!/1!(3-1)! ∙ a² ∙ b¹ + 3!/2!(3-2)! ∙ a¹ ∙ b² + 3!/3!(3 - 3)! ∙ a⁰ ∙ b³
= 3!/1! ∙ 3! ∙ a³ + 3!/1! ∙ 2! ∙ a² b + 3!/2! ∙ 1! ∙ ab² + 3!/3! ∙ 0! ∙ b³
= a³ + 3a² b + 3ab² + b³
= (2x)³ + 3 ∙ (2x)² ∙ (4y) + 3 ∙ (2x) ∙ (4y)² + (4y)³
= 8x³ + 3 ∙ 4x² ∙ 4y + 3 ∙ 2x ∙ 16y² + 64y³
= 8x³ + 48x² y + 96xy² + 64y³

Remark! The terms a and b can be negative as well. All the above operations are carried out in the same way as when both terms are positive but by taking into account the changes determined by any term's sign.