Aptitude Questions on Permutation-Combination

  1. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
    A. 1480
    B. 2520
    C. 5040
    D. 7020
  2. C. 5040
    The Word LOGARITHMS contain 10 letters.
    To find how many 4 letter word we can find from that = 10*9*8*7 = 5040
  3. A delegation of 5 members has to be formed from 3 ladies and 5 gentlemen. In how many ways the delegation can be formed, if 2 particular ladies are always included in the delegation?
    A. 16
    B. 20
    C. 24
    D. 28
  4. B. 20
    There are three ladies and five gentlemen and a committee of 5 members to be formed.
    Number of ways such that two ladies are always included in the committee = 6C3 = (6 * 5 * 4) / 6 = 20
  5. How many 4 digit numbers can be formed using the digits (1, 3, 4, 5, 7, 9) when repetition of digits is not allowed?
    A. 300
    B. 320
    C. 340
    D. 360
  6. D. 360
    The given digits are six.
    The number of 4 digit numbers that can be formed using six digits is 6P4 = 6 * 5 * 4 * 3 = 360
  7. Using all the letters of the word 'THURSDAY', how many different words can be formed?
    A. 7
    B. 7!
    C. 8
    D. 8!
  8. D. 8!
    Total number of letters = 8
    Using these letters the number of 8 letters words formed is 8P8 = 8!
  9. There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?
    A. 280
    B. 330
    C. 380
    D. 430
  10. C. 380
    The total number of stations = 20
    From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in 20P2 ways.
    20P2 = 20 * 19 = 380
  11. The number of new words that can be formed by rearranging the letters of the word 'ALIVE' is:
    A. 117
    B. 118
    C. 119
    D. 120
  12. C. 119
    Number of words which can be formed = 5! - 1 = 120 - 1 = 119
  13. Six points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points?
    A. 30
    B. 32
    C. 34
    D. 36
  14. B. 32
    We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear except those that can be selected out of these 6 points are collinear.
    Hence, the required number of straight lines
    = 11C2 - 6C2 - 5C2 + 1 + 1
    = 55 - 15 - 10 + 2 = 32
  15. A selection is to be made for one post of principal and two posts of vice-principal amongst the six candidates called for the interview only two are eligible for the post of principal while they all are eligible for the post of vice-principal. The number of possible combinations of selectees is:
    A. 4
    B. 12
    C. 18
    D. 20
  16. D. 20
    Total number of ways = 2C1 . 5C2 = 2 * 5!/3!2! = 2 * 10 = 20
  17. In how many ways can the letters of the word 'LEADER' be arranged?
    A. 72
    B. 144
    C. 360
    D. 720
  18. C. 360
    The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
    Required number of ways = 6! / (1!)(2!)(1!)(1!)(1!) = 360
  19. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
    A. 5
    B. 10
    C. 15
    D. 20
  20. D. 20
    Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
    The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
    The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
    Required number of numbers = (1 * 5 * 4) = 20