Quantitative Aptitude: Permutations and Combinations Set 1 (From Basics)

This time in SBI PO Mains 2017 along with probability, a simple question from Permutations and Combinations was also asked. Permutations and Combinations Basic Questions for NICL AO, IBPS RRB PO/Clerk, IBPS PO/Clerk, and other exams.

  1. There are 36 boys and 28 girls in a class. For inter-school competition, 1 boy and 1 girl is to be selected. In how many ways can this selection be done?
    A) 1033
    B) 1008
    C) 1048
    D) 1038
    E) 1030
    View Answer
      Option B
    Solution
    :
    We can choose 1 boy from 36 boys in 36 ways or say 36C1 ways.
    Similarly, we can choose 1 girl from 28 girl in 28 ways
    Since both tasks are to be completed, so number of ways of selection of 1 boy and 1 girl is 36*28 = 1008
  2. How many 5 letter words with or without meaning can be formed from the letters of the word ‘AMONG’, when repetition is allowed and when it is not allowed?
    A) 55, 120
    B) 54, 60
    C) 45, 150
    D) 55, 24
    E) 53, 150
    View Answer
      Option A
    Solution
    :
    We have to make 5 letter words.
    Case 1: Repetition allowed
    Observe the letters to be placed in 5 boxes as:

    Now since repetition is allowed, means any letter can be used up to 5 number of times. So 5 choices for the Box 1, 5 choices for the Box 2,…….., 5 choices for the Box 5. Now since all letters are to be placed, we will use multiplication. So ways = 5*5*5*5*5 = 55
    Case 2: Repetition not allowed
    Observe the letters to be placed in 5 boxes same as above.
    Now since repetition is not allowed, means any letter can be used only once. So 5 choices (A, M, O, N, G) for Box 1.
    Now suppose N was placed in 1st box. Now 4 letters are left for the Box 2 (N cannot be used again).
    Now same way, 3 choices for the Box 3,…….., 1 choices for the Box 5. Again all letters are to be placed, so multiplication.
    So ways = 5*4*3*2*1 = 120  
  3. How many 4 digit numbers can be formed from digits – 1, 2, 4, 6, 7, 8 when repetition is not allowed?
    A) 720
    B) 240
    C) 360
    D) 120
    E) 220
    View Answer
      Option C
    Solution
    :
    Now since 4 digit numbers are to be formed. Use 4 boxes:

    Now total numbers are = 6
    So 6 choices for the Box 1. Now suppose 4 was placed in Box 1. We are left with 5 numbers to be placed in Box 2. So 5 choices for Box 2. Similarly 4 choices for Box 3, and 3 choices for Box 4.
    So total numbers = 6*5*4*3 = 360
  4. How many 6 digit odd numbers can be formed from all digits – 0, 1, 2, 5, 6, 7?
    A) 318
    B) 242
    C) 264
    D) 288
    E) 302
    View Answer
      Option D
    Solution:

    From all digits means repetition is not allowed.
    Take 6 boxes as 6 digit odd numbers are to be formed as in above questions
    Remember 0 cannot be placed in Box 1 because then it will be 5 digit number. When conditions are given, we cannot start will first box. Odd numbers are to be formed means 2, 5, or 7 to be in Box 6.
    So we have 3 choices for Box 6. Suppose 7 is filled in Box 6.
    Now move to Box 1. We have 5 choices left but since 0 cant be placed here. So 4 choices for Box 1 (1, 2, 5, 6).
    Now there is no condition on any other box. So start filling with Box 2. After filling 2 boxes, we have 4 choices left.
    So 4 choices for Box 2, 3 for Box 3, 2 for Box 4, and 1 for Box 5.
    So total numbers =
  5. How many 3 digit even numbers can be formed from digits – 1, 2, 3, 5, 6, 7 when repetition is not allowed?
    A) 60
    B) 50
    C) 30
    D) 40
    E) 20
    View Answer
      Option D
    Solution:

    We have to make 3 digit numbers. so take 3 boxes like above in questions.
    Even number is to be formed – means a condition on last digit. We have 2 even numbers (2 and 6). So 2 choices for Box 3.
    Now there is not any condition on nay box.
    So 5 choices for Box 1, and 4 for Box 2
    So total numbers = 5*4*2 = 40   
  6. How many numbers are there between 100 and 1000 which have exactly one of their digits as 3?
    A) 156
    B) 286
    C) 225
    D) 325
    E) 308
    View Answer
      Option C
    Solution
    :
    Between 100 and 1000 means 3 digit numbers. So make 3 boxes.  Again there is a condition, that 3 should be placed in any of the 3 boxes.
    Remember we are not given about repletion, only numbers are to be formed. So number can be 344 also – repetition allowed.
    Case 1: 3 is in Box 1.
    So 1 choice for Box 1. Now 3 is to be placed exactly once. So out of 10 choices (0-9), 9 choices are left for Box 2 and 3 both.
    So numbers = 1 * 9 * 9 = 81
    Case 2: 3 is in Box 2.
    So 1 choice for Box 2. Now see that 0 cannot be placed in Box 1
    So out of 10 choices (0-9), 8 choices are left for Box 1. and then 9 choices (here 0 is allowed, but not 3) for box 3.
    So numbers = 8 * 1 * 9 = 72
    Case 1: 3 is in Box 3.
    So 1 choice for Box 3. Now see that 0 cannot be placed in Box 1
    So out of 10 choices (0-9), 8 choices are left for Box 1. and then 9 choices (here 0 is allowed, but not 3) for box 2.
    So numbers = 8 * 9 * 1 = 72
    Now cases will get added because any one of the 3 cases will occur
    So total numbers = 81 + 72 + 72 = 225
  7. There are 6 vacant chairs in a row. In how many ways 3 children can sit on these 6 chairs?
    A) 130
    B) 120
    C) 60
    D) 150
    E) 90
    View Answer
      Option B
    Solution
    :
    One child will sit at one chair
    Now since there is a choice of chairs for children, we will make 3 boxes. (Chairs are constant, children can move)
    In Box 1(which is for 1st child) – He has 6 chairs to sit at. So 6 choices.
    Now 1 chair is filled. For 2nd child- 5 choices. And for 3rd child – 4 choices
    So total ways = 6*5*4 = 120    
  8. How many 3-digit even positive numbers can be formed from the digits 0 to 9, when repetition is allowed?
    A) 17/44
    B) 69/220
    C) 47/66
    D) 13/160
    E) 21/170
    View Answer
      Option C
    Solution:
    3 boxes. From 0 to 9 there are 5 numbers which will make number even.
    Case 1: When 0 is that number to be placed in Box 3
    So 1 choice (0) for Box 3, 9 choices (1-9) for Box 1, and 10 choices for Box 2 [repetition allowed]
    So numbers = 9*10*1 = 90
    Case 1: When 0 is not that number to be placed in Box 3
    So 4 choices (2, 4, 6, 8) for Box. Now 0 can’t be placed in Box 1, so 9 (1-9) choices for Box 1. 10 choices for Box 2
    So numbers = 9*10*4 = 360
    Add the cases
    So total numbers = 90+360 = 450
  9. How many numbers greater than 1000 but less than 5000 can be made from digits 0, 1, 3, 5, 6?
    A) 220
    B) 150
    C) 200
    D) 180
    E) 250
    View Answer
    Option E
    Solution:
    When not written about repetition, take it to be allowed.
    1000 to 5000 means 4 digit numbers. So 4 boxes.
    Now 0, 5 and 6 cannot be placed in Box 1. 0 will make it 3-digit number and 5 & 6 will make number greater than 5000.
    So 2 choices (1 and 3) for Box 1, all 5 choices for Box 2, 3, and 4
    So total numbers = 2*5*5*5 = 250
  10. How many 6 letter words can be formed from letters of word ‘QUESTION’ so that the middle 2 places are taken by vowels? (Repetition is not allowed)
    A) 25%
    B) 30%
    C) 33 1/3%
    D) 50%
    E) 22%
    View Answer
      Option D
    Solution:
    QUESTION has 8 letters

    Make 6 boxes for 6 letter words
    Middle 2 positions means 3rd and 4th box
    There are 4 vowels for Box 3, and then 3 vowels for Box 4.
    Now 2 letters are placed out of 8
    So now no condition is left, start with Box 1.
    6 choices for Box 1, 5 choices for Box 2, 4 choices for Box 5, and 3 choices for Box 6.
    So total words

 

 

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24 Thoughts to “Quantitative Aptitude: Permutations and Combinations Set 1 (From Basics)”

    1. M@nish...

      bhai 6th ques samjh aya ho to samjha do ..?

      1. PRADEEP $K$

        its simple
        when unit digit is 7
        then 10 s place should have any digits leaving 7(0—————9)
        => 9 ways
        and after filling 10s place left with hundreth place
        => 8 ways(0—————7)
        tot =9*8 =72
        now when tens digit is 7
        units place hav any digit excpt 7 => 9 ways(0—————9)
        hundredth place after filling unit place will left with 8 ways(0——-7)
        =>8 ways =>totl =9*8=72
        now taking hundreds place
        ones place(0————–9) =>9 ways
        tens place(0—————9)=>9ways
        totl =9*9=81
        =>totl possiblty of 7 no from 100 to 1000 = 81+72+72 =225

        1. M@nish...

          thanks …bhai ap id de sakte ho apni ?

        2. AMAZON

          @Alice help me out here

      2. PRADEEP $K$

        question ko aise smjo hmko 100 to1000 k bch srf asie no find krna jisme 7 ho…:)))

        1. Ambika

          bro mera bhi ques solve kro ek
          see there

          1. PRADEEP $K$

            wwre???

          2. PRADEEP $K$

            are nhi dik rha sster…:))))hhehehe new comment mn h kya que???

          3. Ambika

            grp me dekho bro

          4. PRADEEP $K$

            got ir r ryt???

          5. PRADEEP $K$

            one mins

          6. PRADEEP $K$

            soln shi h …sster…:)

  1. Sneha

    5th ques
    … 6*5* 2 nhi hona chahiye … pls explain

    1. M@nish...

      yes 60 will be ans

      1. Why 6*5 ???

        repetition is not allowed
        6 nos – 1, 2, 3, 5, 6, 7

        Suppose 2 is used at last position
        so now 5 left na

        so 5*4.

        1. M@nish...

          sorry my mistake its 5*4*2= 40

    2. Why 6*5 ???

      repetition is not allowed
      6 nos – 1, 2, 3, 5, 6, 7

      Suppose 2 is used at last position
      so now 5 left na

      so 5*4

  2. golden girl

    thanku mam 🙂

  3. Rise And Shine

    thank u ::)

  4. AmRITa @ bank po

    Subhra mam permu.&combi KA sirf ek hi part he key??

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