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- Arizona State University - Tempe
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- Statistics 421
- Eubank
- Lecture - 01.26.09 - Distinct Permutations and Combinations

Garrett F.

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STP421: Probability January 26, 2009 Page 1 Lecture – 01.26.09 – Distinct Permutations and Combinations Permutation Review Proposition: We have 𝑛 distinct objects. Choose 𝑘 things. The number of permutations: 𝑛 = 𝑃𝑘𝑛 = 𝑛! 𝑛−𝑘 ! Distinct Permutations Proposition: We have 𝑛 objects of 𝑘 distinct types (𝑛1 𝑜𝑓 𝑡𝑦𝑝𝑒 1,𝑛2 𝑜𝑓 𝑡𝑦𝑝𝑒 2,…,𝑛𝑘 𝑜𝑓 𝑡𝑦𝑝𝑒 𝑘) The number of distinct permutations: 𝑛𝐷 = 𝑛!𝑛 1!∗𝑛2!∗…∗𝑛𝑘! Example #1: To elevated the classes linguistic abilities, we’re going to be learning a foreign language: Texan. Texan for “everybody” or “you all” is, of course “yall.” The plural form of “yall” is, therefore, “allyall.” A) How many ways are there to arrange the letters in ALLYALL, assuming that each letter is distinct (ie: we can tell the difference between each of the two A’s and four L’s)? 𝑛 = 7! = 5,040 B) How many distinct permutations are there if each of the letters is not distinct (ie: if we switch the A’s in ALLYALL to form ALLYALL, this is not a new arrangement)? The number permutations of distinct letters (from part A) is equal to the number of distinct permutations multiplied by the permutations of the similar letters. So, # 𝑜𝑓 𝑃𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛𝑠 = # 𝑜𝑓 𝐷𝑖𝑠𝑡𝑖𝑛𝑐𝑡 𝑃𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛𝑠 ∗ 4! ∗ 2! This is the case as, for example, there are 4! ways of arranging the 4 L’s in ALLYALL that were counted as different arrangements in part A. However, rearranging the L’s doesn’t create a distinct permutation, which is what we are counting in part B. 𝑛 = 7 𝑘 = 3 𝑛1(𝐴) = 2 𝑛2 𝐿 = 4 𝑛3 𝑌 = 1 7! = 𝑛 ∗ 4! ∗ 2 → 𝑛 = 7!4!2!1! = 7∗6∗5∗4∗3∗2∗14∗3∗2∗1∗2∗1∗1 = 7∗6∗52 = 105 Example #2: We have the number set {1,1,2,2,2,3,3,3,4,4}. How many distinct ways can we arrange the number set? 𝑛 = 10 𝑘 = 4 𝑛1 = 2 𝑛2 = 3 𝑛3 = 3 𝑛4 = 2 𝑛𝐷 = 10!2!3!3!2! = 25,200 Example #3: There are 7 paintings to be hung in a museum: 4 by the famous artist A, and 3 by the famous artist B. A’s paintings are dated and must be hung in temporal sequence (chronologically from earliest to latest). 𝑛 = 7 𝑘 = 4 𝑛1 𝐴 = 4 𝑛2 𝐵1 = 1 𝑛3 𝐵2 = 1 𝑛4 𝐵3 1 𝑛𝐷 = 7!4!1!1!1! = 7 ∗ 6 ∗ 5 = 210 STP421: Probability January 26, 2009 Page 2 A B A B C Example #4: There are two bars (A and B) downtown as outlined by the grid. Each line segment in the grid represents a block. Happy hour has ended at bar A, but is just beginning at bar B. How many ways are there to get from bar A to bar B? (We are to assume that we aren’t too drunk and that we can make it from A to B in the minimal number of blocks). We need to travel 6 blocks right and 4 blocks up (10 blocks total), regardless of our route. So, 𝑛 = 10 𝑛1 = 6 𝑛2 = 4 𝑛𝐷 = 10!6!4! = 210 Now let’s say that we are aware of a third bar C, that is on the way from bar A to bar B. It’s such a long trip that we’re going to have to stop at bar C to rehydrate. How many ways are there to get from bar A to bar B while making sure to stop at bar C? A to C: 𝑛 = 6 𝑛1 = 4 𝑛2 = 2 𝑛𝐴𝐶 = 6!4!2! = 15 C to B: 𝑛 = 4 𝑛1 = 2 𝑛2 = 2 𝑛𝐶𝐵 = 4!2!2! = 6 A to B through C: 𝑛𝐷 = 15 ∗ 6 = 90 Example #5: We have 10 identical objects (shirts) that we need to put into 4 boxes. A) How many ways can we put the 10 shirts into the 4 boxes if every box must have at least one shirt? The following is one way to arrange the 10 shirts: ∗⊓∗⊓∗ | ∗⊓∗ | ∗⊓∗ | ∗⊓∗⊓∗ ∗ is a shirt, ⊓ is a space between shirts in a box; and a | separates boxes. To arrange these 10 shirts in the 4 boxes is really just calculating the number of distinct permutations of the 6 spaces and 3 box separators. So, 𝑛 = 9 𝑛1 = 6 𝑛2 = 3 𝑛𝐷 = 9!6!3! = 84 General Rule: While placing 𝑛 identical objects into 𝑘 boxes, if none of the boxes can be empty, 𝑛𝐷 = 𝑛−1 ! 𝑘−1 ! 𝑛−𝑘 !. B) How many ways can we put the 10 shirts into the 4 boxes if boxes can be empty? The way to calculate this, is to follow the rule above (part A) and calculate the number of ways as though you have 14 shirts and none of the boxes can be empty. After arranging the 14 shirts into the boxes, you take one shirt out of STP421: Probability January 26, 2009 Page 3 each box (leaving you with 10 shirts and the possibility that boxes may be empty). So, 𝑛 = 13 𝑛1 = 10 𝑛2 = 3 𝑛𝐷 = 13!10!3! = 286 General Rule: While placing 𝑛 identical objects into 𝑘 boxes, if the boxes can be empty, 𝑛𝐷 = 𝑛+𝑘−1 ! 𝑘−1 !𝑛! . Combinations Proposition: We have 𝑛 distinct objects. Choose 𝑘 things, but order doesn’t matter. Number of Combinations: 𝑛 = 𝑛𝑘 = 𝑛!𝑘! 𝑛−𝑘 ! (Binomial Coefficient) Example #1: There’s a new lottery in town called “Randy’s Pick.” When selecting the winner, there are 50 balls, of which 5 are selected without replacement. If order does not matter, how many combinations of 5 balls can be selected? 𝑛𝐶 = 𝐶5 = 50!45!5! = 2,118,76050 Example #2: A committee of 3 people is to be selected from 4 couples. (𝑛 = 8 and 𝑘 = 3). A) How many possible committees can be formed? The order does not matter in selecting a committee of people, so this is a combination problem: 𝑛𝐶 = 𝐶38 = 8!5!3! = 56 B) How many committees of two women and 1 man can be formed? 𝑛 = # 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜𝑠𝑒𝑙𝑒𝑐𝑡 2 𝑤𝑜𝑚𝑒𝑛 # 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜𝑠𝑒𝑙𝑒𝑐𝑡 1 𝑚𝑎𝑛 = 42 41 = 4!2!2! ∗ 4!1!3! = 24 C) How many committees can be formed if spouses can’t serve together? To select the committee, it makes sense to first select 3 couples, and then select one person from each couple. So, 𝑛 = # 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜𝑠𝑒𝑙𝑒𝑐𝑡 3 𝑐𝑜𝑢𝑝𝑙𝑒𝑠 # 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑠𝑒𝑙𝑒𝑐𝑡𝑎 𝑝𝑒𝑟𝑠𝑜𝑛 𝑓𝑟𝑜𝑚 𝑒𝑎𝑐ℎ 𝑐𝑜𝑢𝑝𝑙𝑒 = 43 21 3 = 4! 3!1! 2 3 = 32 Example #3: There are 52 cards in a standard deck of playing cards. How many hands of 13 cards are possible? The order in which you draw the cards does not matter, so this is a combination problem. 𝑛𝐶 = 𝐶1352 = 5213 = 52!13! 52−13 ! = 635,013,559,600 Garrett STP421: Probability

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