Permutations and Combinations

EXERCISE

1. In how many ways can a chairman, vice-chairman and a secretary of a board of 8 members be selected from amongst themselves?

Ans. 336; Hint: 8 * 7 * 6 = 336

2. There are three bus routes between cities A and B; and four bus routes between cities B and C. In how many ways can a person take a round trip by bus from city A to city C via city B, if he does not want to use a bus line more than once ?

Ans. 72; Hint: For A to B and B to C / 3 * 4 For C to B and B to A / 3 * 2

3. There are four routes for travelling from city A to city B. In how many different ways can a person go from city A to city B and return, if for returning (a) any of the routes is taken, (b) the same route is taken, (c) the same route is not taken?

Ans. (a) 16, (b) 4, (c) 12; Hint: (a) 4 * 4 = 16 (b) 4 * 1 = 4(c) * 4 * 3

Solution : Q.1 to Q.3

4. In how many ways çan 5. prizes be given away to 5 boys when (a) each boy is eligible for more than one prize, (b) each boy is eligible for only one prize?

Ans. (a) 3125, (b) 120;

Hint: (a) 5 * 5 * 5 * 5 * 5 = 3125 (b) 5 * 4 * 3 * 2 * 1 = 120

5. Five letters are written which are to be put in five envelopes. In how many ways can this be done?

Ans. 120; Hint: 5 * 4 * 3 * 2 * 1

6. In how many different ways can one try to open a lock having three rings, if each ring contains numbers 0 to 9? If the lock opens by adjusting a particular number, how many unsuccessful attempts to open the lock are possible ?

Ans. 1000, 999; 

Hint: 10 * 10 * 10 = 1000 Successful attempts is only one no.

Solution : Q.4 to Q.6

7. How many different numbers of three digits can be formed with the digits 1, 4, 6, 7, 8, 9 if (a) none of the digits gets repeated, (b) the digits can get repeated?

Ans. (a) 120, (b) 216; Hint: 6 * 5 * 4 = 120 (b) 6 * 6 = 216

8. How many four digit numbers can be formed by using digits 0, 1, 2, 3, 4, 5, 6; such that no digit is used more than once?

Ans. 720

Hint: Being four digit number, it cannot start with 0. Hence, thousand's place can be filled in 6 ways. 6 * 6 * 5 * 4 = 720

9. A cloth merchant has two types of cloth material "cotton" and "terrycot". Each of the cloth materials has two design patterns with five colour shades each. How many choices are open to a customer, who wants to purchase a cloth material from the merchant?

Ans. 20; Hint: 2 * 2 * 5 = 20

10. How many natural numbers not exceeding 5432 can be formed with the digits 2, 3, 4, 5?

Ans. 229; Hint: Numbers less than 5432 have the following possibilities.

Th	H	T	U
Digits 4 or 3 or 2 Digit 5 Digit 5	Any four nos. Digits 3 or 2 Digit 4	Any four nos. Any four nos. Digit 2	Any four nos. Any four nos. Any four nos.

Solution : Q.7 to Q.10

EXERCISE

11. Find the value of:

(i) 6!

(ii) (8!)/(2 x 6!)

(iii) 5! - 2 x 3!

Ans. (i) 720, (ii) 28, (iii) 108

12. Find the value of:

(1) (8!)² /(6!)³

(ii) (8! - 4 x 6!)/(7! - 3 x 6!)

Ans. (1) 196/45 (ii) 13

13. Prove that (n + 1)!/(n - 1)! = n² + n

14. Prove that (20!)/(9! 11!) + (20!)/(10! 10!) = (21!)/(10! 11!)

15. Prove that 11*12*13*.....* 30 = (30!)/(10!)

16. Show that: 30!= 2¹⁵ * 15! (1*3*5.....29).

Hint: 

30!= 30 * 29 * 28 * 27 * 26 *.....*3*2*1 

= (15 * 2) * 29(14 * 2) * 27(13 * 2) *.....*3*2*1 

=[ (15 * 2)(14 * 2) *.....*(2*1)]*(29*27*.....*3*1)

17. Show that: 2!/n! = 2ⁿ * (2n - 1)(2n - 3)(2n - 5) *....*5*3*1.

Hint: 2!=  2n * (2n - 1)(2n - 2)(2n - 3)(2n - 4) *.....*3*2*1 =[ 2n(2n - 2)(2n - 4) *....*2][(2n-1)(2n-3)*....*3*1]

Solution : Q.11 to Q.17

EXERCISE

18. Find the values of: (i) ⁹P₅ (ii) ⁶P₆

Ans. (i) 15120, (ii) 720

19. If  ⁿP₃ :ⁿP₂ = 3:1 find the value of n.

Ans. 5

20. Find n, if ²ⁿP₃ = 100 ⁿP₂

Ans. 13

21. Find the value of n, if (i) ⁿ⁺³P₃ : ⁿ⁺²P₄ = 14 (ii) ^n-1P₃ : ⁿ⁺¹P₃  = 5:12

Ans. (i) 4, (ii) 8

Solution : Q.18 to Q.21

22. Find the value of n, if (i) ²ⁿP₃ = 2 ⁿP4 (ii) ⁿP₃ = ^n-1P₃ + 3 ⁷P₂

Ans. (i) 8, (ii) 8

23. Find the value of r, if

(i) ⁷Pr = 60 ⁷Pr-3

(ii) ⁸P_r = ⁷P₃ + 3 ⁷P₂

(iii) ¹⁶P_{r+5 :} ¹⁴P_r+2  = 480

(iv) ⁹P₅ + 5 ⁹P₄= ¹⁰P_r

(v) ⁵⁶P_r+6 : ⁵⁴P_r+3  = 30800

Ans. (i) 5, (ii) 3, (iii) 10, (iv) 5, (v) 41

Solution : Q.22 to Q.23 (i/ii/iii/iv/v)

24. Find the values of m and n, if m + n*P_{2} = 56 and m - n*P_{2} = 12

Ans. m = 6 n = 2

25. Show that ⁿP₇ = 210  ⁿP5 for n = 20

26. How many different 5 digit numbers can be formed from digits 1, 2, 3, 4, 5, 6, none of the digits being repeated in any one of the numbers so formed?

Ans. 720

27. Find the number of different numbers of six different digits, which can be formed with the digits 0, 1, 3, 5, 8, 9. How many of these have 0 in ten's place?

Ans. 600, 120

Hint: Six digit number should not start with 0 i.e. leading number should not be zero. So, 5 * 5 * 4 * 3 * 2 * 1 = 600

Ten's place should be occupied by zero. Remaining digits to be assigned to other five places (digits should be different) i.e. 1 * 5 * 4 * 3 * 2 * 1 = 120

28. How many words can be formed using all the alphabets of the word "DELHI"?

Ans. 120

Hint: 5! = 120

29. In how many ways can 6 boys and 4 girls be made to stand for a photograph?

Ans. 10!

30. How many words can be formed using four different alphabets of word "COMBINE"?

Ans. 840; 

Hint: ⁷P₄ = 840

31. Find the value of n, if four times the numbers of permutations of n things taken 3 at a time is equal to 5 times the number of permutations of (n - 1) things taken 3 at a time.

Ans. 15; 

Hint: 4 ⁿP₃ = 5 ^n-1P₃ i.e. 4 * (n!)/(n - 3)! = 5 * (n - 1)!/(n - 4)!

32. Six persons, including a couple are to for a photograph in a row so that the couple is always together. Find the number of such arrangements.

Ans. 240; 

Hint: 5! 2! = 240

33. In how many ways can a group of nine boys stand for a photograph in a row, if the leader of the group is to be always at the Centre ?

Ans. 40320; Hint: 1 * 8! = 40320

34. In how many different ways can the letters of the word "THURSDAY" be arranged? How many of these arrangements begin with "T"? How many begin with "7" and end with "U"?

Ans. 40320, 5040, 720; 

Hint: 8! = 40320,  1*7! = 5040, 1 * 1 * 6! = 720

35. How many words can be formed of the letters in a word "FAILURE" such that (i) vowels always come together? (ii) consonants are together and vowel are also together?

Ans. (i) 576, (ii) 288

Hint (i): 4 vowels and 3 consonants.

Vowels can be grouped together to be arranged with and then 4 vowels can be arranged amongst themselves. So, 4!4! = 576

Hint (ii): Two groups i.e. one of vowels and one of consonants can be arranged between themselves, 4 vowels can be arranged among themselves and 3 consonants can be arranged among themselves. So, 2! 4! 3! = 288

36. How many different words containing all the letters of the word "LOGARITHM" be formed? How many of them have (i) none of the consonants together, (ii) vowels occupying odd places, (iii) vowels at the third, fourth and fifth places?

Ans. 362880, (i) 0, (ii) (5!6!)/(2!) (iii) 3! 6!

Hint (i) :  (9!) = 362880

Hint (ii): 5 odd places out of 9 letters can be occupied by 3 vowels. All the remaining places thereafter which can be occupied by consonants. So, (5!6!)/(2!)

37. A number of four different digits is formed using the digits 1, 2, 3, 4, 5, 6, 7, Find (i) how many such numbers can be formed, and (ii) how many of them are greater than 3400?

Ans. (i) 840, (ii) 560

Hint (i): ⁷P₄ = 840

Hint (ii): There can be two situations when the number is greater than 3400.

(a) Thousand's place is filled in by either of the digits 4, 5, 6, 7 and remaining three places by any of the remaining 6 digits.

(b) Thousand's place is filled in by digit 3, hundred's place is filled in by any of the digits 4, 5, 6, 7 and remaining two places by any of the remaining 5 digits.

4* ⁶P₃ + 1 * 4 * ⁵P₂ = 560

38. How many numbers of different digits lying between 100 and 1000 can be formed with the digits 2, 3, 4, 0, 8, 9?

Ans. 100; 

Hint: Number should be three digit number i.e. 5 x ⁵P₂ = 100

39. A football team consists of 5 forwards, 3 half backs, 2 full backs and a goal keeper. If the forwards may interchange positions in forward line, the half backs in their line and the full backs in their line, find the number of different ways in which the team may be arranged in the field.

Ans. 1440; Hint: 5! 3! 2! = 1440

40. In how many ways can 5 books on Accountancy, 6 books on Mathematics and 7 books on Commerce be arranged on a shelf in a row so that all the books on the same subject will always be together ?

Ans. 3! 5! 6! 7!

41. Find the number of numbers less than 1000 and divisible by 5 which can be formed with digits 0 to 9 such that no digit gets repeated.

Ans. 153

Hint: 1 digit no. 1

2 digit nos. (⁹P₁-1)+ ⁹P₁

3 digit nos. (⁹P2-P₁)+ ⁹P2

42. A family consisting of an old man, 6 adults and 4 children, is to be seated in a row for dinner. The children wish to occupy two seats at each end and the old man refuses to have a child on either side of him. In how many ways can the seating arrangement be made for the dinner?

Ans. 86400

Hint: 4 children should occupy two seats on each side first. Old man has five seats available (no child on either side). Remaining seats can be occupied by the adults. i.e. 4! x5x6! = 86400

43. Ten guests are to be seated in a row. Three of them are to be seated together. Of the remaining seven, two do not wish to sit next to each other. Find the number of possible arrangements.

Ans. 181440

Hint: Consider all the number of arrangements with three of them seated together and subtract the number of arrangements where three of them sit together as well as two of them sit together (where really they do not want to sit together). So, 3! 8!7! 3! 2! = 181440.

44. The letters of the word ZENITH are written in all possible orders. How many words are possible, if all these words are written as in a dictionary? What is the rank of the word ZENITH?

Ans. 720, 616

Hint (i): 6! = 720

Hint (ii): All words beginning with E, N, 1, T, H will appear before the word ZENITH. Hence consider the rank of the word starting with Z. The same will be the rank of the word with the first two alphabets ZE. Consider the words starting with ZE and appearing before the word ZENITH i.e. ZEHxxx or ZElxxx. Then, consider the words beginning with ZENHxx and lastly, the word ZENIHT appearing before the word ZENITH. No. of words appearing before the word ZENITH = 5 × 51+2×31+21+1=600+12+2+1=615. Hence rank of the word ZENITH will be 616.

45. A boat has 8 crew members, 3 of whom can row only one side and 2 only on the other side. Find the number of ways in which the crew can be arranged, 

Ans. 72; Hint: 3!2!3! = 72

COMBINATIONS

Definition: Each of the groups or selections which can be made by taking some or all of a number of things without reference to the order of the things in each group is called a combination.

Hence, if n things are given and we have to choose r (<= n) out of them and the order of choosing these things is not important, such a selection is called a combination of n objects taken r at a time and is denoted by ^ n C r or binomial(n,r)

Note: In combinations of n objects taken r at a time, order in which the selection is done is not important.