1. 12/13/2015MATH 106, Section 41 Section 4 Permutations Questions about homework? Submit homework!

2. 12/13/2015MATH 106, Section 42 What are permutations? Many of the problems we will be interested in are concerned with arranging objects in some order. Such ordered arrangements are called permutations. Nine people form a baseball team. How many ways can they “take the field”? (Note: There are nine positions on the field to occupy.) RECIPE: Choose the team member to take first base and then choose the team member to take second base and then choose the team member to take third base and then choose … and then choose the team member to take short stop and then choose the team member to be catcher. 9  8  7  6  5  4  3  2  1 = 362,880 Shortcut notation: 9! (“9 factorial”) #1

3. 12/13/2015MATH 106, Section 43 n factorial In general, n! (“n factorial”) is the product n  (n – 1)  (n – 2)  …  3  2  1 By definition 0! = 1 NOTE: n! = n  (n – 1)! NOTE: n! = n  (n – 1)! So we have … 0! = 1 1! = 1 = 1 2! = 2  1 = 2 3! = 3  2  1 = 6 4! = 4  3  2  1 = 24 … See page 31 of the textbook

4. 12/13/2015MATH 106, Section 44 Let’s work with factorial a little bit … Simplify the following: 5! — 4! 5 6! — 4! 30 8! —— 5! 3! 56 Twelve people form a baseball team. How many ways can they “take the field”? (Note: There are nine positions on the field to occupy.) 12  11  10  9  8  7  6  5  4 = 79,833,600 12! = ——— (12 – 9)! (Same recipe as in #1) #2

5. 12/13/2015MATH 106, Section 45 Seventeen people form a baseball team. How many ways can they “take the field”? (Note: There are nine positions on the field to occupy.) 17  16  15  14  13  12  11  10  9 = 8,821,612,800 17! = ——— (17 – 9)! In general, when we are given n things, but only taking k of them where the order we take them matters, we have what is called the permutation of n things taken k at a time. The number of such permutations and is denoted as P(n, k), and (Same recipe as before!) #3

6. 12/13/2015MATH 106, Section 46 Nine people form a baseball team. How many ways can they “take the field”? (Note: There are nine positions on the field to occupy.) Let’s rework those examples … P(9, 9) Twelve people form a baseball team. How many ways can they “take the field”? (Note: There are nine positions on the field to occupy.) P(12, 9) Seventeen people form a baseball team. How many ways can they “take the field”? (Note: There are nine positions on the field to occupy.) P(17, 9) #1 #2 #3

7. 12/13/2015MATH 106, Section 47 How many ways are there to arrange the letters in “victory”? How many 4-letter arrangements can you form? P(7, 7) = 7! = 5040 How many ways can you choose the class officers of president, vice-president, secretary, and treasurer from a class of 35? P(35, 4) = P(7, 4) = 7! ——— = 840 (7 – 4)! 35! ———– = 1,256,640 (35 – 4)! When we recognize a permutation, we may find it unnecessary to write a recipe. #4 #5 Complete the handout problems we do not finish here in class as part of the homework for next class…

8. 12/13/2015MATH 106, Section 48 Suppose 24 students are to be seated in 24 chairs, and 8 of the students are seniors, 7 are juniors, 5 are sophomores, and 4 are freshmen. How many seating arrangements can you make? How many seating arrangements can you make, if we want all of the seniors first, then all of the juniors, then sophomores, and then freshmen? How many seating arrangements can you make, if we want to keep all students in the same class together, but we don’t care about the order of the classes? P(24, 24) = 24! P(8, 8)  P(7, 7)  P(5, 5)  P(4, 4) = 8!  7!  5!  4! = 585,252,864,000 RECIPE: Seat the seniors and then seat the juniors and then seat the sophomores and then seat the freshmen. #6

9. 12/13/2015MATH 106, Section 49 Suppose 24 students are to be seated in 24 chairs, and 8 of the students are seniors, 7 are juniors, 5 are sophomores, and 4 are freshmen. How many seating arrangements can you make? How many seating arrangements can you make, if we want all of the seniors first, then all of the juniors, then sophomores, and then freshmen? How many seating arrangements can you make, if we want to keep all students in the same class together, but we don’t care about the order of the classes? P(24, 24) = 24! P(8, 8)  P(7, 7)  P(5, 5)  P(4, 4) = 8!  7!  5!  4! = 585,252,864,000 4!  P(8, 8)  P(7, 7)  P(5, 5)  P(4, 4) = 4!  8!  7!  5!  4! = 14,046,068,740,000 RECIPE: Seat the seniors and then seat the juniors and then seat the sophomores and then seat the freshmen. RECIPE: Decide the class seating order and then seat the seniors and then seat the juniors and then seat the sophomores and then seat the freshmen.

10. 12/13/2015MATH 106, Section 410 Consider a standard deck of 52 cards. How many ways can you select 5 cards from the deck and place them in order on a table? How many ordered arrangements can you form selecting 5 cards from the deck, if you allow repeated selections of the same card by placing each selected card back into the deck before the next selection? P(52, 5) = 52! ———– = 311,875,200 (52 – 5)! #7 NOTE: A standard deck of 52 cards consists of 13 cards in each of 4 suits named clubs, diamonds, hearts, and spades. The 13 cards in each suit consists of denominations ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king.

11. 12/13/2015MATH 106, Section 411 Consider a standard deck of 52 cards. How many ways can you select 5 cards from the deck and place them in order on a table? How many ordered arrangements can you form selecting 5 cards from the deck, if you allow repeated selections of the same card by placing each selected card back into the deck before the next selection? P(52, 5) = 52  52  52  52  52 = 52! ———– = 311,875,200 (52 – 5)! 52 5 = 380,204,032 #7

12. 12/13/2015MATH 106, Section 412 How many 8-symbol computer passwords are there if you can have letters or digits with no repetitions? letters or digits with no repetitions but it must start with a letter? letters or digits with repetitions allowed? letters or digits with repetitions allowed but it must start with a letter? P(36, 8) = 36  36  …  36  36 = 36! ———– (36 – 8)! 36 8 26  P(35, 7) = 35! 26  ———– (35 – 7)! 26  36  …  36  36 = 26  36 7 #8 Just set up these calculations without actually calculating.

13. 12/13/2015MATH 106, Section 413 How many computer passwords are there which can have at most 4 symbols if you can have letters or digits but no repetitions? P(36, 1) + P(36, 2) + P(36, 3) + P(36, 4) = 36 + 1260 + 42,840 + 1,413,720 = 1,457,856 You have 6 men and 6 women to seat in a row of 12 seats. How many seating arrangements are possible if you alternate men and women? 2  P(6,6)  P(6,6) =1,036,800 RECIPE: Choose a computer word with one symbol or choose a computer word with two symbols or choose a computer word with three symbols or choose a computer word with four symbols. RECIPE: Choose which sex will sit in the first seat and then assign seats for men and then assign seats for women. #9

14. 12/13/2015MATH 106, Section 414 Signals are made by running three colored flags up a mast. There is an unlimited supply of flags in seven different colors. How many signals are possible with no repetitions of color allowed? How many signals are possible with repetitions of color allowed? How many signals are possible if no two adjacent flags are allowed to be the same color? P(7, 3) = 7  7! ———– = 210 (7 – 3)! 7  7  7 = 7 3 = 343 6  6 = 252 #10

15. 12/13/2015MATH 106, Section 415 Homework Hints: In Section 4 Homework Problem #2, In Section 4 Homework Problem #14, In Section 4 Homework Problem #15, you can write the answer in scientific notation if you would prefer not to write out the 19 digits required for the answer. note that you are working with a total of 4 + 3 + 5 different books. note that you must decide on the order of the subjects first, after which you must decide the order within each subject.