11.7 The Fundamental Counting Principle and Permutations 727 Example 7 could also be worked using the fundamental counting principle because we are discussing an ordered arrangement (a permutation) that is done without replacement . For the first date scheduled, there are 10 possible outcomes. For the second date selected, there are 9 possible outcomes. By continuing this process we would determine that the number of possible outcomes for the 6 different trips is 10 9 8 7 6 5 151,200. ⋅ ⋅ ⋅ ⋅ ⋅ = We have worked permutation problems (selecting and arranging, without replacement, r items out of n distinct items) by using the fundamental counting principle and using the permutation formula. When you are given a permutation problem, unless specified by your instructor, you may use either technique to determine its solution. calculate the number of different permutations of selecting and arranging the dates for 6 out of 10 possible routes. P 10! (10 6)! 10! 4! 10 9 8 7 6 5 4! 4! 151,200 10 6 = − = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = There are 151,200 different ways that 6 routes can be selected and scheduled from the 10 possible routes. 7 Now try Exercise 45 TECHNOLOGY TIP Evaluating Permutations Most scientific and all graphing calculators can evaluate permutations. While the commands on these calculators may vary depending on the model, look for a key or function that has the notation nPr. If you cannot locate this key or function, conduct an Internet search with your calculator model and the word permutation . For example, if you search for “Casio FX-82MS permutation” you will be directed to several websites including online videos that can assist you. Also, there are many apps for smartphones and tablets that can be used to calculate permutations. To evaluate P 10 6 on a TI-84 Plus calculator, enter the number 10. Then press the MATH key. Use the right arrow key to scroll to PRB, then scroll to nPr. Press the ENTER key. Next, press 6. Press the ENTER key again. The calculator will display 151200, which agrees with our answer in Example 7. The screenshots below show the calculation depending on the version of the TI-84 Plus. Permutations of Duplicate Items So far, all the examples we have discussed in this section have involved arrangements with distinct items. Now we will consider permutation problems in which some of the items to be arranged are duplicates. For example, the name BOB contains three letters, of which the two Bs are duplicates. How many permutations of the letters in the name
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