Quantitative Aptitude: Permutations and Combinations Set 2

  1. A student is to answer 10 out of 15 questions in an examination such that he must choose at least 5 from the first six questions. The number of choices available to him is
    820
    910
    745
    882
    None
    Option D
    Solution:
    No of choices=6c5.9c5 +6c6.9c4
    =6*[(9*8*7*6)/(4*3*2*1)] + 1*[(9*8*7*6)/(4*3*2*1)] =6*126 +1*126
    =756+126
    =882.
  2. The number of ways in which a team of ten players can be selected from 20 players including 2 of them and excluding 4 of them is
    14c8
    20c10
    20c8
    16c10
    None
    Option A
    Solution:
    2 players already included.
    Now 8 players have to selected from 14 players.
    14c8 ways.
  3. In how many ways 3 prizes can be given away to 10 boys when each boy is eligible for any of the prizes?
    2000
    1000
    1800
    1500
    None
    Option B
    Solution:
    No of ways=10*10*10=1000ways.
  4. 3 boys and 2 girls are to be seated in a row in such a way that two girls are always together. In how many different ways can they be seated?
    42
    48
    54
    66
    None
    Option B
    Solution:
    2Girls always together then consider them as one.
    Now there are 4 members.
    And that two girls can sit in 2! Ways
    Then No of ways=4!*2!
    =4*3*2*1*2
    =48ways.
  5. How many four digit number can be formed whose unit digits is always zero and repetition of the digit is not allowed?
    385
    422
    486
    504
    None
    Option D
    Solution:
    In a 4 digit no unit digit is fixed ie 0.
    Then in remaining 3 digit is from no 1-9
    Then no of ways=9*8*7*1= 504.
  6. From 6 men and 3 women, a committee of 5 member is to be formed. In how many ways, if the committee has to include at least one women?
    90
    105
    120
    85
    None
    Option C
    Solution:
    Ways= (1 women *4men) +(2 women*3men)+ (3 women *2men)
    =3c1*6c4 +3c2*6c3 + 3c3*6c2
    =3*15 + 3*20 +1*15
    =45+60+15
    =120.
  7. In how many ways can a leap year have 53 Sundays?
    3
    6
    5
    2
    None
    Option D
    Solution:
    In a leap year there are 366 days i.e. 52 weeks + 2 extra days.
    Then 53 Sundays can occur in only 2 ways.
    i.e. (Saturday and Sunday) or (Sunday and Monday).
    no of ways =2.
  8. There are five trains running between station A and B. In how many ways can a man go from A to B and return by a different train?
    30
    25
    20
    15
    None
    Option C
    Solution:
    A Man go from A to B in 5 ways by 5 train available.
    He return from B to A in 4 ways by the remaining 4 train only.
    Then no of ways=5*4=20ways.
  9. How many words can be formed by re-arranging the letters of the word CURRENT such that C and T occupy the first and last position respectively?
    60
    120
    40
    80
    None
    Option A
    Solution:
    First and last position stable.
    C_ _ _ _ _T
    Remaining URREN must be change
    Ie
    5!/2!
    =120/2=60.
  10. How many different straight lines can be formed by joining 12 different points on a plane of which 4 are collinear and the rest are non-collinear?
    61
    48
    56
    74
    None
    Option A
    Solution:
    Total no of lines formed by 12 points=12c2
    No of lines forms by 4 collinear points=4c2
    Then Required no of lines=12c2 – 4c2 +1
    =66-6+1
    =61.


 

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