1. Permutation: An arrangement of r objects from n objects, the order of which is important. The possible number of such arrangements is denoted by n P r 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 n C r 5B.1 Permutations and Combinations 23/10/20151

2. 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. Permutation (جایگشت), Combination (ترکیب) 23/10/20152

3. 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 5B.1 Permutations and Combinations 23/10/20153

4. 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: a.the digits cannot be repeated? الف: تکرار مجاز نیست. b.the digits can be used more than once? ب: تکرار مجاز است a. 10 P 4 = 10!/(10-4)! = 10!/6! = 5040 or 10*9*8*7 b.10*10*10*10 = 10000 5B.1 Permutations and Combinations 23/10/20154

5. 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 Combination (ترکیب) 23/10/20155

6. 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. 54 C 6 = 54!/6!(54-6)! = 54!/(6!*48!) = 25,827,165 دقت دارید که ترتیب چیدمان کارتها مهم است. 5B.1 Permutations and Combinations 23/10/20156

7. Twenty people are chosen at random. a.What is the probability that none have the same birthday? b.What is the probability that at least 2 have the same birthday? Graph y = 1 – (365 n P r x)/(365^x) with a window of [0,47], [0,1] Birthday Problem 5B.1 Permutations and Combinations 23/10/20157

8. 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 4 C 4 = 4!/4!(4-4)! = 1 Note 0! = 1 With 3 toppings 4 C 3 = 4!/3!(4-3)! = 4 With 2 toppings 4 C 2 = 4!/2!(4-2)! = 6 With 1 topping 4 C 1 = 4!/1!(4-1)! = 4 1 + 4 + 6 + 4 = 15 different types of pizzas 5B.1 Permutations and Combinations 23/10/20158

9. How many ways are there arrange the following letters? HALEY REITER DEREK 5! = 120 6!/(2!2!) = 180 5!/(2!) = 60 5B.1 Permutations and Combinations 23/10/20159

10. What is the probability of getting four of a kind with 5 cards dealt from a standard deck of 52? 5B.1 Permutations and Combinations 23/10/201510