      Combinations In Probability           The concept of combinations is easy to understand. A combination describes picking an unordered subgroup from a larger group of items. A permutation is similar, but it is used when the order within the subgroup matters. For example, imagine that you have three children. At work you were given free movie passes. This should make your children very happy, except there is a problem. You were only given two passes. One of your children must stay at home while the other two go. The order in which you pick the lucky two doesn't matter; both will get to see a movie. This is an example of a combination.

Now imagine that you arrived late to the cinema, probably because you spent much time consoling your third child. There are only two movies playing, and each screening has only one seat left. You decide that the first child can choose the movie, and the second must see the other show. Now the order of the two children matters, and we have converted this example into a permutation.

Of course, your children want to know how lucky, or unlucky, they were. They want to know the number of possible ways that the two moviegoers could have been picked. Let's assume that some people have left the theater, and the two kids can see whatever they want. We are back to a combination problem and need to count the number of combinations. If the children are Jane, John and Sue, then these are the possibilities:

Jane, John
Jane, Sue
John, Sue

There are three possible combinations. Whether this information actually makes the children any happier, is a question left to the reader.

Although it's fine to count the number of combinations by drawing out the possibilities in this way, it only works with small numbers. What if this is the old woman who lives in a shoe? What if there are nine children and two movie tickets? It isn't feasible to draw out all the ways to choose two out of nine. She needs the general formula:

C(n,r) = n! / ( (r)!(n-r)! )

The exclamation point (!) is a factorial, "n" is the total number of children and "r" is the number of moviegoers. So, "n" is nine, and "r" is two.

The number of ways to choose two moviegoers out of nine children is:

C(9,2)
= 9! / ( 2! * (9-2)! )
= 9! / (2! * 7!)

To calculate a factorial, just multiply the factorial number with all of the positive integers below it.

9! = 9*8*7*6*5*4*3*2*1
2! = 2*1
7! = 7*6*5*4*3*2*1

The number of combinations for the mother who lives in a shoe is:

(9*8*7*6*5*4*3*2*1) / ( (2*1) * (7*6*5*4*3*2*1) )
= 362,880/(2 * 5040)
= 362,880/10080
= 36

For those with an algebra background, you'll notice that the formula can be simplified by canceling out terms between the numerator and denominator. If not, then just find the calculator on your smartphone!