How many ways are there to select ‘k’ outcomes from ‘n’ possible outcomes, without concern for the order (or sequence) of the selection? The answer lies in the **binomial coefficient**, which also answers several related questions in combinatorics (the study of countable objects) and algebra.

## Let’s Keep the Discussion of Binomial Coefficients “Positive”

This article discusses only *non*-negative integer values of ‘k’ and ‘n,’ so “n ≥ k ≥ zero” and “both ‘n’ and ‘k’ are integers.”

Although the “Gamma function”, ‘Γ(x)’, extends the factorial function beyond the non-negtive integers, it will *not* be discussed here.

## Introducing the Binomial Coefficient

When this factorial formula: “C(n, k) = n! / ( k! * (n – k)! )” is read aloud as “n choose k,” its meaning is to “choose any ‘k’ of ‘n’ items in any order.” What does this mean in a real example from combinatorics, the mathematics of counting?

Let’s say we have the four aces from a standard deck of playing cards, and want to select a pair. The formula for “4 choose 2” provides the calculation.

- C(4, 2) = 4! / ( 2! * 2! ) = (4*3*2*1)/((2*1)*(2*1)) = 6

Let’s count out the possibilities. Let **A**={c, d, h, s} represent the aces of clubs, diamonds, hearts and spades. The “2-subsets of **A**” is the set of pairs from **A**, disregarding order. This set is {(c,d), (c,h), (c,s), (d,h), (d,s), (h,s)}. Note that there are six unique pairs.

In general, “n choose k” gives the cardinality, or size, of the set of “k-subsets” that can be chosen from a set with ‘n’ elements.

Try to solve this exercise: Each side in the game of chess has bishops, a King, knights, pawns, a Queen and rooks. Therefore there are 6 types of pieces. Let P={b, K, k, p, Q, r}. In how many ways can one select any two types from this set?

## Why Use the Word “Binomial?”

The word “**binomial**” relates to the two terms inside the brackets in the construction “(x + y)^n” for a non-negative integer ‘n.’

In the expansion of “(x + y)^n = (x + y)*(x + y)*…*(x + y),” there are terms such as “m*(x^j * y^(n-j))” for every “j from zero to n.” The integer ‘m’ in this expansion is the “**coefficient.**” Here is an example:

- (x + y)^2
- = (x + y)*(x + y)
- = x*x + x*y + y*x + y*y
- =
*1**x^2 +*2**x*y +*1**y*2″

The above coeficients are ‘*1*‘, ‘*2*‘ and ‘*1*‘.

The “Binomial Theorem” states that in any expansion of “(x + y)^n,” the coefficient for “x^j * y^(n-j)” is C(n, j).

## The Symmetry Rule of Binomial Coefficients

One unusual but obvious fact about the binomial coefficient is the symmetry rule. “C(n, k) = C(n, n-k).”

The simple algebraic proof uses the commutative property of multiplication.

- C(n, n-k) = n! / ( (n – k)! * (n – (n – k))! )
- = n! / ( (n – k)! * k! ) , by simplifying “n – (n – k)” = “n – n + k” = ‘k’
- = n! / ( k! * (n – k)! )
- = C(n, k) , as required

**Click to Read Page Two: Pascal’s Rule for Binomial Coefficients**

tusegiu says

and so that

papasu says

sure this can be as shown prof dr mircea orasanu in more cases