Quantitative Aptitude-More exercise questions

Permutation and Combination-Exercise Questions

|   32999

1. Evaluate 50!

                47!

a. 102500

b. 112584

c. 117600

d. 118450

 

2. Find the value of 85P3 .

a. 565350

b. 595650

c. 535950

d. 565350
 

3. Find the value of (20C18)*(20C20)

a. 400

b. 380

c. 360

d. 350

 

4. How many words with or without meaning, can be formed by using all the letters of the word, ‘ORANGE’, using each letter exactly once?

a. 700

b. 720

c. 750

d. 800
 

5. There are 28 stations between Ernakulam and Chennai. How many second class tickets have to be printed, so that a passenger can travel from one station to any other station?

a. 800

b. 820

c. 850

d. 870

 

6. In how many ways can the letters of the word, ‘TECHNOLOGY’ be arranged?

a. 1804400

b. 1814400

c. 1714400

d. 1704400

 

7. A bag contains 2 yellow balls, 3 white balls and 5 red balls. In how many ways can two balls be drawn from the bag?

a. 2C2

b.10C2

c.8C2

d.5C2

 

8. In how many ways can the letters of the word, ‘LANGUAGE’ be arranged in such a way that the vowels always come together?

a. 600

b. 700

c. 720

d. 750

 

9. In how many ways can the letters of the word, ‘KEYBOARD’ be arranged in such a way that the vowels always come together?

a. 4250

b. 4520

c. 4320

d. 4230

 

10. In how many ways can a team 16 be chosen out of a batch of 20 players?

a. 4845

b. 6852

c. 3125

d. 5846

 

11. How many ways can the letters of the word, ‘MACHINE’ be arranged so that the vowels may occupy only the odd positions?

a. 210

b. 576

c.144

d. 456

 

12. From a group of 5men and 4 women, 3 persons are to be selected to form a committee so that at least 2 men are there are on the committee. In how many ways can it be done?

a. 20

b. 50

c. 65

d. 86            

                                             

13. In how many ways can a committee consisting of 4 men and 5 women be formed from a group of 7 men and 9 women?

a. 7C4  9C5

b. 4C7  5C9

c. 7C5  9C4

d. 9C4  7C5

 

14. In how many ways can 5 boys and 3 girls sit around a table in such a way that no two girls sit together?

a. 1000

b. 1400

c.1440

d. 1800

 

Directions for questions 15 to 16:  Refer the data below and answer the questions below:

A letter lock has 3 rings each containing 6 letters.

 

15. What is the maximum number of false trials that can be made before the lock is opened?

a. 3*26C6

b. (26C6)3

c. 26C6. 3!

d. 215

 

16. How many such three letter passwords can exist?

a. 216

b. 26C6 *3

c. (26C6)3

d. (26C6)3 *63

 

17. How many different words can be formed from the word DAUGHTER so that ending and beginning letters are consonants?

a. 7200

b. 14400

c. 360

d. 1440

 

18. Out of 6 consonants and 3 vowels, how many words of 4 consonants and 2 vowels can be formed?

a. 1050

b. 25200

c. 32400

d. 5800

 

19. A box contains 3 white balls, 4 black balls and 5 yellow balls. In how many ways can 4 balls be drawn from the box, if at least one yellow ball is to be included in the draw?

a. 652

b. 547

c.425

d. 356

 

20. In how many ways can 22 books on English and 20 books on Hindi be placed in a row on a shelf so that two books on Hindi may not be together?

a. 4586

b. 5896

c. 2415

d. 1771

 

Answer & Explanations

 

1. Evaluate 50! = 50*49*48* (47!) = 50*49*48 = 117600

                  47!              47!

2. 85P3 = 85!    = 85!  = 85*84*83*82! = 85*84*83 = 595650

             (85-3)!  82!              82!

3. 20C20 = 1

 

    (20C2)*(20C20) = 20! * 1= 20*19*18! = 20*19*1= 380

                             18!              18!

4. Exp: The word ‘ORANGE’ contains 6 different letters.

    Therefore, Required number of words= Number of arrangement of 6 letters, taken all at a time

    =6P6 = 6!= 6*5*4*3*2*1= 720

5. Exp: The total number of stations= 30

    From 30 Stations we have to choose any two stations and the direction of travel (Ernakulam to Chennai

    is different from Chennai to Ernakulam) in 30P2 ways. 

 

    30P2= 30*29= 870

6. Exp: The word ‘TECHNOLOGY’ contains 10 letters namely 2O, 1T, 1E, 1C, 1H, 1N, 1L, 1G, 1Y.

    Therefore, Required number of ways=                                                       10!     10!                                                                                   (2!)(1!) (1!) (1!) (1!) (1!) (1!) (1!) (1!)         2!

                                            = 10*9* 8*7*6*5*4*3*2*1 = 1814400

                                                            2*1

7. Exp: Total number of balls= 2+3+5 = 10

     2 balls can be drawn from 10 balls in 10C2

8. Exp: In the word ‘LANGUAGE’ we treat the vowels AUAE as one letter. Thus, we have LNGG (AUAE).

    This we have 5 letters of which G occurs 2 times and the rest are different.

     Number of ways arranging these letters=   5! =5*4*3= 60

                                                                 2!     

     Now, 4 letters of which A occurs 2 times and the rest are different, can be arranged in 4! = 4*3=12.

                                                                                                                             2!    

     Therefore,  Required number of ways= 60*12= 720

9. Exp: In the word ‘KEYBOARD’ we treat the vowels EOA as one letter. Thus, we have KYBRD (EOA).

 

    Thus we have 6 letters can be arranged in 6! = 720 ways

    The vowels (EOA) can be arranged among themselves in 3! = 6 ways

    Therefore, Required number of ways= (720*6)= 4320

10. Exp: Required number of ways= 20C16= 20C(20-16) = 20C4

                                                  = 20*19*18*17  = 4845.        

                                                        4*3*2*1

11. Exp: In the word ‘MACHINE’ 3 vowels and 4 consonants.

      v      v      v      v

      __ _ __ _ __ _ __

    Now, 3 vowels can be placed at any of 3 places, out of which 4 marked 1,3,5,7.

    Number of ways arranging the vowels= 4P3 = (4*3*2) =24

    Also, 4 consonants at the remaining 4 positions may be arranged in = 4P4 = 4!= 24 ways.

    Therefore, Required number of ways= (24*24) =576.

12. We have (2men and 1 woman) or (3men only)

    Therefore, Required number of ways= (5C2*4C1) + (5C3)

                                              =   5*4   * 4   +  5C2

                                                   2*1

                                              =  10*4 + 10

                                              = 40 +10 =50

13. Exp: Group consisting of 7 men and 9 women

     4 men can be selected from 7 men in 7C4 ways

     5 women can be selected from 9 women in 9C5 ways

    Therefore, Total number of ways= 7C4  9C5

14. Exp: The 5 boys can be seated around a table in 4! Ways. In between them there are 5 places.

     The 3 girls can be placed in the 5 places in 5P3 ways.

    Therefore, Required number of ways= 4!* 5P3

                                              =24*60=1440

15. Exp:

        

     Maximum possible permutation of letters= 6*6*6= 216

     Out of 216 different permutations only 1 is correct.

     Maximum number of false trials= 216-1= 215

16. Exp: 1st ring: 6 out of 26 alphabets can be selected in 26C6 ways.

     And is for 2nd and 3rd ring.

     Also, these 3 set of 6 letters can be arranged amongst themselves in 63 ways.

     Hence, total number of 3 letter passwords = 26C* 26C*26C6 *63 ways.

17. Exp: Here total letters are 8,3 vowels and 5 consonants. Here 2 consonants can be chosen in 5C2 ways
     and these 2 consonants can be put it in 2! Ways. The remaining 6 letters can be arranged in 6! Ways.

     The words beginning and ending letters with consonant = 5C2 *2! *6! = 14400

18. Number of ways of selecting (4 consonants out of 6) and (2 vowels out of 3)

                                              =  6C4 * 3C2

                                              = 6C2 * 3C1

                                              = 6*5  *3

                                                  2*1

                                              =  15*3= 45

     Number of groups, each having 4 consonants and 2 vowels= 45.

     Each group consists of 6 letters.

     Number of ways of arranging 6 letters among themselves

                                             = 6! = (6*5*4*3*2*1)= 720

    Therefore, Required number of words= 45*720= 32400

19. Exp: We may have  (1 yellow and 3 others) or  (2 yellow and 2 others) or (3 yellow and 1 others) or (4
      yellow).

    Therefore, Required number of ways= (4C1*8C3)+ (4C2*8C2)+ (4C3*8C1)+ (4C4)

                                             

                                             =  4* 8*7*6    +   4*3  *  8*7   +  (4C1*8) +1

                                                       3*2*1         2*1     2*1   

                                             = 224+168+32+1= 425.

20. Exp: In order that two books on Hindi are never together, we must place all these books as under:

     H E H E H E H….. H E H

     Where H denotes the position of Hindi book and E that of English book.

     Since there are 22 books on English, the number of places marked E are 23.

     Now, 20 places out of 23 can be chosen in 23C20 = 23C3 = 23*22*21

                                                                                        3*2*1    

                                                                              = 1771 ways.

    Hence the number of ways = 1771 ways  

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