PREVNEXTSUBMITMenuResourcesGlossaryNotesClear and return to MenuSearch ResultsFilterSlide TextNotesSearch in:Check to includeSearch...TermsDefinitiondefinitionresourcessearchpauseplayreplaysubmitnextpreviousvolumepermutations: arrangements of distinct objects example 1 evaluate the following expressions (a). p(8, 3) (b). p(6, 5) (c). p(9, 6) example 1 evaluate the following expressions the number of arrangements of 3 out of a set of 8 objects is 336 the number of arrangements of 5 out of a set of 6 objects is 720 the number of arrangements of 6 out of a set of 9 objects is 504 using the online calculator we can use the online calculator to determine the permutations using the built-in function. click on the image on the right hand side to see the demonstration using example 1. example 2 in a card game, each player is dealt a face down "reserve" of 13 cards that can be turned up and used one by one during the game. how many different sequences of reserve cards could a player have? solution: in this example, you are taking 13 cards from a deck of 52. therefore: there are approximately 3.95 x1021 different sequences of reserve cards a player could have. example 3 how many different "words" of size 5 can you make from the letters of the alphabets? solution: this is an example of arranging 5 items from a set of 26. therefore, it is a permutation. therefore: total number of possible "words" are 7893600. example 4 how many three-letter permutations can be formed from the letters of the word dinosaur? solution: this is an example of arranging 3 letters from a set of 8. therefore, it is a permutation. there are 336 ways. another way of looking at this example is that the first letter has to come from 8 choices, second from 7, and third from 6. therefore, applying the fundamental counting principle, we have 8 x 7 x 6 = 336 ways. example 5 find the number of permutations of the letter of the word diploma for which the letter l remains in the middle position. solution: this is an example in which we calculate the number of permutations possible when specific conditions must be met. the technique of filling boxes, as used in the introduction, is helpful in dealing with variations of straightforward permutation situations. therefore we use seven boxes with l in the middle position as follows: example 5 then the middle box can be filled only with the letter l, that is, in one way, while the other six letters can be arranged freely. so there are 1 x p(6, 6) = 6! = 720 ways of arranging letters of diploma with the l in the middle. example 6 at a used car lot, six cars for sale are to be parked side-by-side. in how many different ways can this be done if the one (and only) car with a sunroof must be at the right end of the line? solution: the various orders of the cars could appeal in different ways to customers, so the order is significant. the number of different ways the cars could be parked is: 1 x p(5, 5) = 1 x 5! = 120 ways example 7 at a used car lot, six cars for sale are to be parked side-by-side. in how many different ways can this be done if the three black cars must be together? solution: the three black cars are considered as one unit., so there are four items to be arranged. however, there are 3! ways of arranging the black cars within the one unit. so, there are: 3! x p(4, 4) = 3! x 4! = 144 ways of parking the cars with the three black cars together. example 8 how many six-letter words can be formed from the letters of travel? solution: if any of the six letters can be used. so, there are: p(6, 6) = 6! = 720 ways now let's suppose the first letter must be a, then we have: 1 x p(5, 5) = 1 x 5! = 120 ways end of slide show click here to go to the beginning