Permutations on the ACT and GRE Today I would like to talk about some more advanced methods for solving Counting Problems. The content presented here and in tomorrow’s post is meant for more advanced students whose current ACT math score is at least a 25 or whose GRE score is at least a 148. Last week we discussed some more basic methods which can be viewed by clicking the following link: Listing and the Counting Principle. Please quickly review yesterday’s post before going on to the content below. Five minutes of review will really solidify the concepts from that post and will make the information in this post easier to learn. Factorials First let’s discuss factorials. The factorial of a positive integer n, written n!, is the product of all positive integers less than or equal to n. n! = 1·2·3…n 0! is defined to be 1, so that n! is defined for all nonnegative integers n. The definition for 0! may seem strange to many of you. Mathematicians have made this definition so that the formula n! = n(n – 1)! will be true for all positive integers n. Note that with this definition 1! = 1·0! Permutations Now a permutation is just an arrangement of elements from a set. The number of permutations of n things taken r at a time is For example, the number of permutations of {1, 2, 3} taken 2 at a time is These permutations are 12, 21, 13, 31, 23, and 32. The good news is that on the ACT you do not need to know the permutation formula. You can do this computation very quickly on your graphing calculator. To compute 3P2, type 3 into your calculator, then in the Math menu scroll over to Prb and select nPr (or press 2). Then type 2 and press Enter. You will get an answer of 6. On the GRE, of course, you would not be able to use this calculator feature. As always, let’s internalize these principles by putting them into practice. Here is a simple counting problem that can be solved using permutations. Example 1 Four different books are to be stacked in a pile. In how many different orders can the books be placed on the stack? Solution by listing: Okay, now you could imagine stacking the books in different ways. Let’s try this by listing the different permutations. To distinguish the books let’s make one red, one blue, one yellow, and one green. Let’s abbreviate each book’s color by using its first letter. rbyg rbgy rybg rygb rgby rgyb bryg brgy byrg bygr bgry bgyr yrbg yrgb ybrg ybgr ygrb ygbr grby gryb gbry gbyr gyrb gybr We now see that there are 24 arrangements. The thing is that on a standardized test I would rather not take the time to write out each permutation. There is a chance, under the time pressure of the test, that an error could be made or too much time could be spent checking that I have not duplicated any of the permutations. Here are two more solutions that are much more efficient. First we can use the counting principle that we reviewed in last week’s post: Solution using the counting principle: There are 4 possible books for the bottom of the stack. After placing the first book, there are 3 possible books that can go on top of the bottom book, then 2 books for the next position, and then 1 book for the top of the stack. Using the counting principle we get a solution of (4)(3)(2)(1) = 24 arrangements. Or we can solve this problem even faster by using our new knowledge of permutations: Solution using permutations: There are 4 books, and we are arranging all 4 of them. Therefore there are 4P4 = 4! = (1)(2)(3)(4) = 24 arrangements. I think you would agree that the last solution is much faster than the original solution by listing. Example 2 Three light bulbs are placed into three different lamps. How many different arrangements are possible for three light bulbs of different colors – one red, one green, and one yellow? Solution by listing: Again, we can list all the possibilities, abbreviating each color by using the first letter. rgy ryg gry gyr yrg ygr We can easily see that there are 6 arrangements. And again, we can also use the counting principle: Solution using the counting principle: There are 3 possible lamps to place the red bulb in. After placing the red bulb, there are 2 lamps to place the green bulb in. Finally, there is 1 lamp left to place the yellow bulb in. By the counting principle we get (3)(2)(1) = 6 arrangements. And finally we can solve this problem very quickly using our knowledge of permutations: Solution using permutations: There are 3 light bulbs, and we are arranging all 3 of them. So the number of arrangements is 3P3 = 3! = 1·2·3 = 6. Where To Find More Practice Problems If you are preparing for the ACT, GRE or any other standardized test that has counting problems, you may want to take a look at the Get 800 collection of test prep books. Click on the picture below for more information. And if you liked this article, please share it with your Facebook friends: Comments comments