This Edited Entry requires at least a basic knowledge of algebra.

How can we find out the chances of winning the National Lottery each week? If we were to pick five cards at random, what is the probability of getting a run? There exists a single theorem that can be used to work these probabilities out. This is called binomial theorem and it has its origins in the most abstract of algebra.

In algebra, it is often neccessary to expand terms of the form (x+y)^{n} in order to facilitate cancellation. It can be seen by inspection that this will form a series^{1} of terms ranging from x^{n} to y^{n} with every other power of x and y featuring in between. The general problem for expansion, therefore, is to find how many of each power combination of x and y there are. These are called the *co-efficients* of the series.

**Pascal's Triangle**

For small (up to about n = 12) integral powers, n, there is a quick method to determine the co-efficients of the terms. Although references are found in Chinese mathematical texts to the technique, it is generally attributed to the French mathematician, Blaise Pascal. Pascal used the method extensively in his work on probability where binomial co-efficients are used to calculate combinations of items. It is this feature we can use since the number of terms that have a particular power of x and power of y is the number of combinations of that power in the expansion.

Pascal's triangle has many properties which are fascinating to study and play with. We will be using just one here to generate the whole triangle. Any term in a particular row is the sum of the two terms above it in the row above.

Starting from n=0 with the first term 1, the triangle extends down as follows:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

... and so on.

This allows us to derive such facts as (x+y)^{2} = x^{2} + 2xy + y^{2} and (x+y)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}. However, by about n=12 this method of generating the triangle becomes cumbersome. It also suffers problems that it isn't generalised in algebra and it can only cope with positive integral values n. We need a more general method. Let's look at the link between Pascal's triangle and probability.

**Combinatorics**

Consider a bag with n items. From that bag we wish to pick r items. How many combinations of r items are there where order doesn't matter? (That is, abcd say is the same as acdb.) Let us think about the first item. How many ways can we pick that one item? Clearly there must be n items to pick from so n ways. How many ways for the second item? There is now one less item in the bag so the number of ways is (n-1). This gives us n.(n-1) ways of picking the first two items. How many ways are there to pick the whole bag? If we continue this method down we get the expression n.(n-1).(n-2). ... .3.2.1. This is given the shorthand of *n!* or *n factorial*.

We now have an expression to find the number of ways to pick all the items out of a bag. We wish to eliminate all the terms from r to zero to find the number of ways to pick r from a bag of n. Since a factorial is a product, we can do this by dividing by the terms we wish to eliminate. These are all the terms from (n-r) down. Hence the expression becomes:

Number of ways of picking r from n = n! / (n-r)!

The number of ways of picking r from n *where order matters* is called the number of *permuations* of r from n. This is denoted by ^{n}P_{r}.

It can be shown that the number of orders of r items is r!. Therefore, to find the number of *combinations* where order doesn't matter we must divide ^{n}P_{r} by r!. This result is denoted by ^{n}C_{r} or (^[n]_[r]). It is this last form which is the general notation for binomial co-efficients.

**What Else Is it Used for?**

The story goes that two gamblers were playing some game of chance which was interupted. Not wanting to discard the pot, they decided to split the winnings according to the probability of winning at that point. However, they couldn't work it out. In frustration they turned to Blaise Pascal to work out the probabilities of each of them winning. This gamble gone wrong started Pascal in a discussion with Pierre de Fermat that culminated in the foundations of Probability Theory, Game Theory and Statistics.

Earlier in the article the National Lottery was mentioned. Let us work out just how likely getting the jackpot is. We know that in the lottery 49 balls are used out of which six are chosen randomly. Assuming we only play one row, what is the probability of winning? To win we need to get all six balls (however, the order isn't important). The probability of any event, P(X), is equal to the number of successes over the total range of results. We only can have one successful combination and yet there are ^{49}C_{6} possible combinations of six balls out of 49. This lands us with a P(X) of 1 in 13,983,816. This is less than the chance of being hit and killed by an asteroid each week.

**How Does it Apply to a Binomial Series?**

The above result gives us two ways of working out the binomial expansion. The first works only for positive integral n and is as follows;

(1) (x + y)^{n} = /sigma _[r=0] ^[r=n] (^[n]_[r]) . x^{n - r} . y^{r} where n /ismember /gothic{Z^{+}}

The second is more involved but does work for negative and fractional n^{2} and is as follows;

(2) (x + y)^{n} = x^{n} + nx^{n-1}y + /frac{n.(n-1)}{2!} x^{n-2}y^{2} + ...