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So there is 1 possibility involving 3 dogs, 2 possibilities involving 2 dogs, and 3 possibilities involving 1 dog Since all possibilities involving dogs are exhausted, it should be easier to pick out all the remaining combinations involving the remaining two variables: CCC CCM CMM MMM Can you think of any other possibilities Let s hope not, because there are no others The answer is 10, which is choice D Note that this system would have worked just as well if you d started with monkeys or cats The trick is to start on a system that will inevitably exhaust all possibilities Brute force has its limitations, though Say you encounter the following question: A sock drawer contains seven socks, each one a different solid color The colors of the socks are red, orange, yellow, green, blue, indigo, and violet If the socks are laid next to each other, how many different arrangements of socks could be made A 24 B 240 C 1,024 D 2,520 E 5,040 First, understand what is being asked You re looking for permutations of seven variables let s call them R, O, Y, G, B, I, and V in groupings of seven, with no repeat variables in each grouping, and order in this case does matter (ie, ROYGBIV is not the same as ROYGBVI) You could write out every possible permutation, starting like this: ROYGBIV ROYGBVI ROYGVIB ROYGVBI ROYGIVB ROYGIBV ROYBGIV ROYBGVI ROYBVIG ROYBVGI ROYBIVG ROYBIGV ROYVBIG and so on

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documentwrite(lastModgetMonth() + "-" + lastModgetDate());

As you can see, with some problems, the brute force method could take a very long time On this particular problem, it might take you the whole test, as you might be able to guess by looking at the scope of the available answers It is in problems like this one that combinatorics formulas come in handy

The key concept in combinatoric math is the factorial The factorial of an integer n is the successive product of n times all the positive integers smaller than n, right down to 1 In practice, that means that n! = n (n 1) (n 2) (n 3) 1

9 Finish the date with another + - + and lastModgetFullYear( )

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Actually, the 1 is wholly unnecessary, but you get the point The factorial function blows up rather quickly, so you ll not be asked to calculate a factorial much larger than these: 0! = 1 (it is simply defined that way) 1! = 1 2! = 2 1 = 2 3! = 3 2 1 = 6 4! = 4 3 2 1 = 24 5! = 5 4 3 2 1 = 120 6! = 6 5 4 3 2 1 = 720 7! = 7 6 5 4 3 2 1 = 5,040 Before we move on to the formulas, let s consider how the concept of factorials alone could be used to solve the sock problem above Suppose the question asked: In how many different ways could you arrange a single red sock in a sock drawer if there are exactly 7 spaces for socks The answer, clearly, is 7 Now, what if the question asked: In how many different ways could you arrange a single red sock and a single orange sock in a sock drawer if there are exactly 7 spaces for socks The answer is slightly more complicated this time There are 7 ways you could arrange the red sock Once the red sock is put in the drawer, there are now only 6 remaining spots in which to put the orange sock, so the total possible number of arrangements is 7 6 = 42 Now what if the question asked: In how many different ways could you arrange a single red sock, a single orange sock, and a single yellow sock in a sock drawer if there are exactly 7 spaces for socks This time, we have 7 ways to put in the red sock, 6 places left to put in the orange sock, and then 5 remaining places left to put in the yellow sock So the total number of possible arrangements is 7 6 5 = 210 possible arrangements of these 3 socks in 7 spaces.

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