Binomial Coefficient

Definition of the binomial coefficient

Let n be a positive integer and k a positive integer less than or equal to n, then the binomial coefficient is the number of subsets of k elements choosen from the n elements (abbréviated to "n choose k"). The binomial coefficient is given as,

`([n], [k]) = C_n^k = frac {n!}{k!(n-k)`

n! denotes factorial n,

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

Newton binomial and binomial coefficient

Binomial coefficients are used to calculate the coefficients of a polynomial raised to a power n.

Example: calculate the coefficient of $x^4y^2$ in the expansion of $(x + y) ^6$

$\text {coef} (x^4y^2) =\dbinom {6} {4} =\dfrac {6!} {4! \, 2!} = 15$,

We'd calculate all the coefficients rather than expanding $(x + y) ^6$ which can be tedious!

Other coefficients:

$\text {coef} (x^6) =\dbinom {6} {6} =\dfrac {6!} {6! \, 0!} = 1$
$\text {coef} (x^5y) =\dbinom {6} {5} =\dfrac {6!} {5! \, 1!} = 6$
$\text {coef} (x^4y^2) =\dbinom {6} {4} =\dfrac {6!} {4! \, 2!} = 15$ (already calculated above)
$\text {coef} (x^3y^3) =\dbinom {6} {3} =\dfrac {6!} {3! \, 3!} = 20$
$\text {coef} (x^2y^4) =\dbinom {6} {2} =\dfrac {6!} {2! \, 4!} = 15$
$\text {coef} (xy^5) =\dbinom {6} {1} =\dfrac {6!} {1! \, 5!} = 6$
$\text {coef} (y^6) =\dbinom {6} {0} =\dfrac {6!} {0! \, 6!} = 1$

Therefore,

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Notice that some coefficients are equal : $x^2y^4$ and $x^4y^2$ , $xy^5$ and $x^5y$. This symmetry is easily deduced from the binomial coefficient formula. This simplifies the calculations. Applied to $(x+y) ^8$ polynom, we get,
$\text {coef} (x^8) = \text {coef} (y^8) = 1$

For the other coefficients, we only calculate the coefficients of $xy^7$,$x^2y^6$,$x^3y^5$ and $x^4y^4$. The other coefficients are deduced symetrically.

See also

Combination