4_6Permutations.notebook. April 22, 2013

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5 Permutations A Permutation is an ordered arrangement of objects selected from a set. P(n,r) represents the number of permutations possible in which "r" objects from a set of "n" different objects are arranged. P(n,r) can also be written as n P r We looked at situations like this before. We can make a tree diagram or we can learn to calculate using permutations. Example Children frequently play with interlocking plastic blocks. These blocks come in six different colours (eg. red, blue, white, green, grey and black). Suppose you have one block of each colour. a) How many different six block sequences can be formed? b) How many different two block sequences can be formed? 5

6 Factorials To calculate Permutations, we use something called a "Factorial". n! (read as n factorial) is calculated as n! = n x (n 1) x (n 2) x... x 2 x 1 For example, 4! = 4 x 3 x 2 x 1 = 24 Look for the factorial button on your calculator. It saves time (more efficient) to type 5! rather than 5 x 4 x 3 x 2 x 1 also note that 0! = 1 Factorials in permutations: 6

7 Back to our example Children frequently play with interlocking plastic blocks. These blocks come in six different colours (eg. red, blue, white, green, grey and black). Suppose you have one block of each colour. a) How many different six block sequences can be formed? This is a permutation problem because order is important in our solution. (we cannot chose a colour more than once) There are 6 possibilities for the first block. There are 5 possibilities for the second block. There are 4 possibilities for the third block... etc. Our solution would calculate as 6 x 5 x 4 x 3 x 2 x 1 or using permutations P(6, 6) = 6! / (6 6)! = 6! / 0! = 720 / 1 = 720 (wow, I'm glad we didn't try to draw a tree diagram!) Did you notice? P(6,6) = 6! b) How many different two block sequences can be formed? Again here order is important (meaning we cannot choose a colour more than once) P(6,2) = 6! / (6 2)! = (6x5x4x3x2x1) / (4x3x2x1) = 6x5 = 30 7

8 How many 4 letter "words" can be made using the letters QUIP? 4 x 3 x 2 x 1 = 4! How many 4 letter "words" can be made using the letters QUIP if Q and U must be together? 3 x 2 x 1 = 3! NOTE: Always deal with restrictions first and "blocked" objects are considered one item but if we consider that QU and UQ are different, yet still satisfy the restriction that Q and U must be together then our answer is 3! x 2 = 12 possible "words" 8

9 How many different 6 letter "words" can be made from the word OTTAWA? At first glance we might think there are 6! different "words" But there are multiples of some letters. We are not able to distinguish between one "T" and the other "T" Since there are 2! ways to arrange the Ts and 2! ways to arrange the As, then there will be 2! x 2! duplicates. To account for this over counting we simply divide 6! / (2! x 2!) or 6! / (2!2!) 9

10 Sometimes it is more efficient to use the COMPLEMENT (All possible arrangements) (arrangements you don't want) = (arrangements you do want) Brooklyn has 6 different coloured sweaters to pile up in her drawers. How many ways can she stack them so that the red sweater is not on the bottom? Using Direct Reasoning: TIP Deal wit the restriction first. The red sweater cannot be on the bottom. So there are 5 other sweaters that can be on the bottom. Since there are 5 other sweaters that can be arranged in any order we know the number of arrangements is 5! Answer: 5 x 5! = Using Indirect Reasoning (the complement): All Red sweater on bottom = Red sweater NOT on the bottom If there were no restrictions we would see 6! possible arrangements If the Red sweater was on the bottom there would be only 5! arrangements Red sweater NOT on the bottom = 6! 5! = 10

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15 PRACTICE: p255 # 1, 4, 7, 12 15

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