A Binomial Coefficient represented as nCk has the following value.
Factorial Formula:
nCk = n! / [(k !) × (n – k)!]The above formula can also be re-written as the below Multiplicative Formula
nCk = [n x (n - 1) x (n - 2) x .... x (n - k + 1)] / [k x (k-1) x (k -2) x ... x 1]Examples
5C3 = (5 x 4 x 3) / (2 x 1) = 60 / 2 = 30
5C1 = (5) / (1) = 5
Binomial Coefficients appear in different branches of math.
- Count the number of ways to select 2 items from A, B, C, D. We get 6 ways, (A, B), (A, C), (A, D), (B, C), (B, D) and (C, D). This is an example of Binomial Coefficient. In general choosing k items from n has nCk different ways.
- Find the value of (x + y)n. For example (x + y)2 = x2 + 2xy + y2 and (x + y)3 = x3 + 3xy + 3xy2 + y3. The coefficients of x and y are Binomial Coefficients. In general (x + y)n = Sum(nCk x n-k yk) for k = 0 to n.
- Consider a situation where you play against a strong opponent in a game and your probability of losing is 60% in every game. What is the probability that you win 2 games out of 3 games. The answer is (0.4)*(0.6)*(0.4) + (0.6)*(0.4)*(0.4) + (0.4)*(0.4)*(0.6) 0.288. Find the number probability of k successes in n trials where every trial is binary (has p probability of success and (1 - p) probability of failure). The answer to this is nCk (1 - P)n-k pk . If we apply the formula for the question discussed, we get the same result 3×(0.40)2×(0.60)1
Interesting Mathematical Properties.
- Symmetry Rule : The value of nCk is equal to nC(n-k). For example 5C3 has value (5 x 4 x 3) / (2 x 1) which is same as 5C2 = (5 x 4 x 3 x 2) / (3 x 2 x 1). The reason behind this symmetry is simple, the number of ways of choosing k items is equal to the number of ways of not choosing n-k items. Note that when we choose k items, we basically do not choose the remaining n-k items.
- Recursive Formula : nCk = (n-1)C(k-1) + (n-1)Ck with base cases as nC0 = nCn = 1. This formula can be proved using the factorial formula discussed in the beginning.
- Sum : Sum of nCk for k = 0 to n is 2n . One way to look at the working of this formula is, for every element, there are two sets of selections. In one set, the element is present and in the other set, not present. So if we consider all elements, we get 2 x 2 x ... 2 (n times) which is 2n
- Sum of kCr for k = 0 to n is (n+1)C(r+1). For example, n = 4 and r = 2, Sum(kC2) for k = 0 to 4 is 5C3
- Catalan Numbers appear in a lot of real world problems. For example. it is the count of valid expressions with n pairs of brackets. For example, if n = 2, then there are two valid expressions ()() and (()). The value of the n-th Catalan can be written in the form of Binomial Coefficient. ( 1 / (n + 1))2nCn
- Pascal Triangle contains Binomial Coefficients that appear in (x + y)0, (x + y)1, (x + y)2, (x + y)3, etc. These coefficients also written in a way that the values of a row are computed using the previous row (using the recursive formula of the Binomial Coefficients). Pascal Triangle has many interesting properties like sum of the n-th row is 2n
