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5B.1 Permutations and Combinations

5B.1 Permutations and Combinations

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5B.1 Permutations and Combinations

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  1. 5B.1 Permutations and Combinations Permutation: An arrangement of r objects from n objects, the order of which is important. The possible number of such arrangements is denoted by nPr Combination: An arrangement of r objects from n objects, the order of which is not important. The possible number of such arrangements is denoted by nCr

  2. Permutation (جایگشت), Combination (ترکیب) جایگشت (Permutation): یک چیدمان r شی از n شی که ترتیب چیدمان مهم است. مثلا AB با BA دو چیدمان متفاوت است. این نوع چیدمان با نمایش داده میشود.

  3. 5B.1 Permutations and Combinations How many ways are there for a Minnesota Twins manager to make out a batting order of 9 players out of a group of 12? به چند طریق میتوان یک تیم نه نفره از مجموع 12 نفر انتخاب کرد؟ 12*11*10*9*8*7*6*5*4 = 79,833,600 or = 79,833,600

  4. 5B.1 Permutations and Combinations • How many 4 digit ATM codes are possible • using the digits 0 – 9 if: • به چند طریق میتوان یک عدد 4 رقمی با استفاده از اعداد 0 – 9 نوشت: • the digits cannot be repeated? الف: تکرار مجاز نیست. • the digits can be used more than once? ب: تکرار مجاز است • 10P4 = 10!/(10-4)! = 10!/6! = 5040 or 10*9*8*7 • 10*10*10*10 = 10000

  5. Combination (ترکیب) ترکیب (Combination): یک چیدمان r شی از n شی که ترتیب چیدمان مهم نیست. مثلا AB با BA یک چیدمان یکسان است. این نوع چیدمان با نمایش داده میشود. معلوم است که همواره

  6. 5B.1 Permutations and Combinations How many ways are there to win at Megabucks? به چند طریق میتوان 6 کارت از 54 کارت را انتخاب کرد؟ Megabucks involves matching 6 numbers out of 54. 54C6 = 54!/6!(54-6)! = 54!/(6!*48!) = 25,827,165 دقت دارید که ترتیب چیدمان کارتها مهم است.

  7. 5B.1 Permutations and Combinations Birthday Problem Twenty people are chosen at random. • What is the probability that none have the same birthday? • What is the probability that at least 2 have the same birthday? Graph y = 1 – (365 nPr x)/(365^x) with a window of [0,47], [0,1]

  8. 5B.1 Permutations and Combinations How many ways are there to make a pizza with toppings of cheese, pepperoni, onions, and sausage if at least one topping is used? • With 4 toppings 4C4 = 4!/4!(4-4)! = 1 Note 0! = 1 • With 3 toppings 4C3 = 4!/3!(4-3)! = 4 • With 2 toppings 4C2 = 4!/2!(4-2)! = 6 • With 1 topping 4C1 = 4!/1!(4-1)! = 4 1 + 4 + 6 + 4 = 15 different types of pizzas

  9. 5B.1 Permutations and Combinations How many ways are there arrange the following letters? 5! = 120 • HALEY • REITER • DEREK 6!/(2!2!) = 180 5!/(2!) = 60

  10. 5B.1 Permutations and Combinations What is the probability of getting four of a kind with 5 cards dealt from a standard deck of 52?