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						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!
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