Permutations

1. Permutations

2. Permutations Permutation:The number of ways in which a subset of objects can be selected from a given set of objects, where order is important. Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is important? (AB, AC, BC, BA, CA, CB)

3. Definition: • A permutation of a set of set of objects is an ordering of the objects in a row. • A permutation of n objects taken r at a time (also called r-permutation of n objects) in an ordered selection or arrangement of r of the objects. P(n,r) = n! / (n-r)! • Ex: set of elements a,b,c Has 6 permutations a,b,c a,c,b b,c,a b,a,c c,a,b c,b,a

4. 2- permutations of { a, b } with unlimited repetitions are : • aa ,ab, ba, bb

5. Ex: set of elements a,a,a,b,c. Has 3- permutations : aaa, aab ,aba, baa, aac,aca, caa,abc, acb,bac, bca,cab, cba.

6. Suppose selection are to be made from the four objects {a,b,c,d} then the 2- permutations without repetitions ? ab,ba ac,ca, ad,da bc,cb, bd,db, cd,dc. Total : 12

7. Suppose selection are to be made from the four objects {a,b,c,d} then the 3- permutations without repetitions ? abc, acb, bac, bca, cab, cba, abd, adb, bad, bda, dab, dba, acd, adc, cad, cda, dac, dca, bcd, bdc, cbd, cdb, dbc, dcb Total : 24

8. Enumerating r-permutations without repetitions • P(n , r)=n(n-1)(n-2).........(n-r+1)=n!/(n-r)! • Since there are n distinct objects, • the first position of r-permutation can be filled in n ways, • the second position of r-permutation can be filled in n-1ways, • so on and rth position can be filled in n-r+1 ways. • From the definition of factorials • P(n , r)= n!/(n-r)!

9. When r = n P(n , n) = n!/0! = n! Corollary: There are n ! permutations of n distinct objects. There are 3! Permutations of {a,b,c} There are 4! Permutations of {a,b,c,d}

10. The number of 2-permutations of {a, b, c, d, e} is • Solution: • n=5 r=2 • C(5,2) = • n! 5! • (n-r)! (5-2)!

11. Examples: • How many ways can the letters in the word C O M P U T E R be arranged in a row? COMPUTER = 8 letters 8 letters can be arranged in 8! ways

12. How many ways can the letters in the word C O M P U T E R be arranged if the letters CO must remain next to each other as a unit. Sol:COMPUTER = CO is treated as 1 unit Then, number of characters= 7 7 letters can be arranged in 7! Ways Note: don’t consider OC. Because CO is consider as a unit.

13. Evaluate p(5,2) n! n=5, r=2 (n-r)! 5! (5-2)!

14. How many 4-permutations are there of a set of seven objects. n! N=7, r=4 (n-r)! 7! (7-4)!

15. How many 5-permutations are there of a set of five objects. n=5, r=5 n! 5! (n-r)! (5-5)!

16. How many different ways can the letters of the word BYTES be chosen and written in a row? Sol: BYTES - 5 letters 5 letters can be chosen in 5! ways

17. How many different ways this can be done if the first letter must be B. Sol: BYTES - 1st letter B is fixed, remaining 4 letters left to choose. So, 4 letters can be chosen in 4! ways

18. r-permutations example • How many ways are there for 5 people in this class to give presentations? (Assume that there are 80 students in class) • There are 80 students in the class • P(80,5) = 80!/(80-5)! • =80*79*78*77*76 = 2,884,801,920 • Note that the order they go in does matter in this example!

19. In how many ways can 7 women and 3 men be arranged in a row if the 3 men must always stand next to each other? • Sol: WW WWWWWMMM (7women+3men) • 7 women can be arranged in 7! Ways. • 3 men can be arranged in 3! Ways. • But 3 men must always together so consider as a unit. • Number of elements to arrange 8 in 8!ways • Finally 8! * 3!

20. In how many ways can the letters of the English language be arranged so that there are exactly 5 letters between the letters a and b? Sol: a _ _ _ _ _ b Number of alphabets =26 • Already starting ending characters are given. • Number of characters left are= 26-2=24. • Number of ways to choose 5 letters is P(24,5). • There are 2 ways to place a and b. • and then 20! ways to arrange any 7-letter word treated as one unit along with the remaining 19 letters • Number of ways to arrange 7 letters word is Finally 2 * 20! * P(24,5)

21. There are (n-1)! Permutations of n distinct objects in a circle. • Example: In how many ways can 5 children arrnage themselevs in a ring. in (n-1)! Ways. n=5 So finally 4! ways

22. Combinations

23. Combinations • Let n and r be non negative integers with r<=n. An r-combination of a set of n elements is a subset of r of the n elements. The symbol C(n , r) Which is read as “n choose r” denotes No. of subsets of size r that can be chosen from a set of n elements.

24. Enumerating r-combinations without repetitions • C(n , r) = P(n , r)/r! = n!/(n-r)! r!

25. Let S={Andhra, Bangalore,chennai,Delhi} 3-combination of S is • C(4,3) = 4 • {Bangalore,chennai,Delhi} • {Andhra, chennai, Delhi} • {Andhra, Bangalore,Delhi} • {Andhra, Bangalore,chennai}

26. In an ordered selection it is not only what elements are chosen but which the order in which they are chosen matters. • In an unordered selection it is only the identity of chosen elements that matters.

27. In how many ways can a hand of 5 cards be selected from a deck of 52 cards Sol: n=52, r=5 C(n,r) = n! 52! (n-r)! r! (52-5)! 5!

28. How many 5-card consists only of hearts? Sol: n=13, r=5 There are only 13 hearts Selecting 5 cards from 13 cards C(n,r) = n! 13! (n-r)! r! (13-5)! 5!

29. How many 5-card hands consists of cards from a single suit. Sol: • there are 4 suits( diamonds, clubs, spades, hearts) • Each suit consists of 13 cards. • 1 Suit selection out 4 suits in 4 ways • Selecting 5 cards from 13 n=13, r=5 • Selecting 5 cards from 13 cards • C(n,r) = n! 4* (13!) (n-r)! r! (13-5)! 5!

30. How many 5-card hands have 2 clubs and 3 hearts? Sol: there are 4 suits( diamonds, clubs, spades, hearts) Each suit consists of 13 cards. 2 cards from clubs(13) C(13,2) ways 3 cards from hearts(13) C(13,3) ways Selecting 5 cards n=13, r=5 C(n,r) = C(13,2) * C(13,3)

31. How many 5-card hands have 2 aces and 3 kings? Sol: there are 4 aces and 4 kings 2 cards from aces (4) C(4,2) ways 3 cards from kings(4) C(4,3) ways Selecting 5 cards n=5, in C(4,2) * C(4,3)

32. In how many ways can a committee of 5 be chosen from 9 people? Sol: selecting 5 people from 9 people n=9 r=5 Chosen in C(9,5) ways

33. How many committees of 5 or more can be chosen from 9 people? Sol: selecting 5 and more i.e. 5 or 6 or 7 or 8 or 9 people can be chosen in C(9,5)+C(9,6)+C(9,7)+C(9,8)+C(9,9)

34. In how many ways an eight letter word with 3 different vowels and 5 different consonants can be framed? • There are 21 consonants and 5 vowels in the English alphabet. • _ _ _ _ _ _ _ _ 8 letter word, • 3 vowels can be chosen from 5 is C(5,3) • 5 consonants can be chosen from 21 is C(21,5) • 8 letters can be arranged in 8! ways • Answer= 8! *C(5,3) * C(21,5)

35. A)how many such words can be formed? • C(5,3) * C(21,5) • B)contain a ‘a’ fixed. • only 2 vowels need to choose from 4 vowels in C(4,2) ways. • 5 consonants can be chosen from 21 is C(21,5) • 8 letters can be arranged in 8! ways • Answer: 8 * C(4,2) * C(21,5)

36. C)contain letters a and b • ‘a’ fixed out of 5 vowels and b is fixed out of 21 consonants • only 2 vowels need to choose from 4 vowels in C(4,2) ways. • Only 4 consonants can be chosen from 20 is C(20,4) • 8 letters can be arranged in 8! ways • Answer: 8 * C(4,2) * C(20,4)

37. D)contain letters b and c • b and c 2 letters are fixed out of 21 consonants • 3 vowels choose from 5 vowels in C(5,3) ways. • Only 3 consonants can be chosen from 19 is C(19,3) • 8 letters can be arranged in 8! ways • Answer: 8 * C(5,3) * C(19,3)

38. E)contain letter a ,b, and c • a is fixed out of 5 vowels • b and c 2 letters are fixed out of 21 consonants • Choose only 2 vowels from 4 vowels in C(4,2) ways. • Only 3 consonants can be chosen from 19 is C(19,3) • 8 letters can be arranged in 8! ways • Answer: 8 * C(4,2) * C(19,3)

39. F)begin with a and end with b • a’ fixed out of 5 vowels • and b is fixed out of 21 consonants • only 2 vowels need to choose from 4 vowels in C(4,2) ways. • Only 4 consonants can be chosen from 20 is C(20,4) • 6 letters can be arranged in 6! Ways (first & last letters already given) • Answer: 6 * C(4,2) * C(20,4)

40. G)begin with b and end with c b is fixed out of 21 consonants Only 3vowels need to choose from 4 vowels in C(5,3) ways. Only 3 consonants can be chosen from 19 is C(19,3) 6 letters can be arranged in 6! Ways (first & last letters already given) Answer: 6 * C(5,3) * C(19,3)

41. There are 30females and 35 males in junior class while there are 25 females 20 males in senior class. In how many ways can a committee of 10 be chosen so that there are exactly 5 females and 3 juniors on the committee? Sol: juniorsseniors Female male female male 0 3 5 2 C(30,0)*C(35,3)*C(25,5)*C(20,2) 1 2 4 3 C(30,1)*C(35,2)*C(25,4)*C(20,3) 2 1 3 4 C(30,2)*C(35,1)*C(25,3)*C(20,4) 3 0 2 5 C(30,3)*C(35,0)*C(25,2)*C(20,5) FINALLY C(30,0)*C(35,3)*C(25,5)*C(20,2) + C(30,1)*C(35,2)*C(25,4)*C(20,3) + C(30,2)*C(35,1)*C(25,3)*C(20,4) + C(30,3)*C(35,0)*C(25,2)*C(20,5)

42. PERMUTATIONS AND COMBINATIONS WITH REPETITIONS

43. PERMUTATIONS WITH REPETITIONS • Let U(n,r) denote the number of r permutations of n objects with unlimited repetitions is U(n,r) = nr Each of the r positions can be filled in ‘n’ ways. Object 1 is repeated in n times Object 2 is repeated in n times Object 3 is repeated in n times : : Object r is repeated in n times Then number of ways = nr

44. There are 25 true or false questions on an examination .How many different ways can a student do the examination if he or she can also choose to leave the answer blank? • Sol: each question have 3 options( T / F / blank) • 1st question answer - in 3 ways (T / F / blank) – 3 • 2nd question answer - in 3 ways (T / F / blank) – 3 • 3rd question answer - in 3 ways (T / F / blank) – 3 • : • : • rth question answer – in 3 ways (T / F / blank) – 3 • Number of different ways = 3 * 3* 3* ….. *3 • = 325

45. The results of 50 foot ball games(win ,lose, tie) are to be predicted .How many different forecasts can contain exactly 28 correct results? Sol: • Number of games got the correct result= C(50,28) • Remaining games left= 50-28=22 • 22 games got the wrong result. • Each game has 2 options to get the wrong results • 1st game has 2 options • 2nd game has 2 options • : • : • 22nd game has 2 options • Wrong results= 222 • Finally different forecasts= C(50,28) * 222

46. Enumerating r-combinations with unlimited repetitions Let the distinct objects be a1, a2,……. an so that the selections are made from {∞. a1, ∞. a1,… ∞. an }. • Any r-combination will be of the form {x1. a1,x2. a2, .... ,xn. an }.where x1……..xn are the repetition numbers and x1+ x2+ x3........+ xn=r

47. V(n,r)=the number of r-combinations of n distinct objects with unlimited repetitions = the number of non negative integral solutions to x1+ x2+ x3........+xn=r =the number of distributing r similar balls into n numbered boxes. =the number of binary numbers with n-1 ones and r zeros =C(n-1+r,r)=C(n-1+r,n-1) =C(n+r-1)!/[r!(n-1)!]

48. The number of 4-combinations of {∞. a1, ∞. a2,… ∞. a5 } Sol: n=5 r=4 C(n-1+r,r) = C(5-1+4,4) = C(8,4) • The number of 3-combinations of 5 objects with unlimited repetitions is n=5 r=3 C(n-1+r,r) = C(5-1+3,3) = C(7,3) • The number of ways of placing 10 similar balls into six numbered boxes is n=6 r=10 C(n-1+r,r) = C(6-1+10,10) = C(15,10) • The number of non integral solutions to x1+x2+x3+x4+x5=30 is n=5 r=30 C(n-1+r,r) = C(5-1+30,30) = C(34,30)

49. The number of binary numbers with ten 1’s and five 0’s is n-1 = 10 r=5 C(n-1+r,r) = C(10+5,5) = C(15,5) • How many different outcomes are possible by tossing 10 similar coins? each coin has two options (heads/ tails) n=2 r=10 C(n-1+r,r) = C(2-1+10,10) = C(11,10) • Tossing 20 similar dice? each dice has six options (1,2,3,4,5,6) n=6 r=20 C(n-1+r,r) = C(6-1+20,20) = C(25,20)

50. 20 similar books placed on 5 different shelves n = 5 r=20 C(n-1+r,r) = C(5-1+20,20) = C(24,20) • Out of large supply of pennies, nickles, dimes, and quarters in how many ways can 10 coins be selected? n= 4(different types of coins) r=10 C(n-1+r,r) = C(4-1+10,10) = C(13,10) • How many ways are there to fill a box with a dozen doughnuts chosen from 8 varieties of doughnuts? n= 8 r= 12 C(n-1+r,r) = C(8-1+12,12) = C(19,12)