H2 Maths combination

• quailmaster

  19 Feb 10, 11:10

  How many even numbers greater than 5000 can be formed with the digits 1,4,5,8,9 if each digit can only be used once?

  First digit 5, 8 or 9                    last digit: 4 or 8

  Case1: 1^st digit 8, last digit 4  3 X 2 = 6

  Case 2: 1^st digit 9, last digit 4  3 X 2 = 6

  Case 3: 1^st digit 9, last digit 8  3 X 2 = 6

  Case 4: 1^st digit 5 , last digit 4 or 8   3 X 2 X 2 = 12 

   

  The answer given is 78

   

   

  Mod: Edited to remove the html tags :)

   

• JDGao

  19 Feb 10, 12:45

  u miss out the number "1"

  so u miss out all the 5 digitis and 4digits with "1"

• Forbiddensinner

  19 Feb 10, 13:24

  Originally posted by quailmaster:

  How many even numbers greater than 5000 can be formed with the digits 1,4,5,8,9 if each digit can only be used once?

  First digit 5, 8 or 9                    last digit: 4 or 8

  Case1: 1^st digit 8, last digit 4  3 X 2 = 6

  Case 2: 1^st digit 9, last digit 4  3 X 2 = 6

  Case 3: 1^st digit 9, last digit 8  3 X 2 = 6

  Case 4: 1^st digit 5 , last digit 4 or 8   3 X 2 X 2 = 12 

  The answer given is 78

  The answer given is incorrect.

  If last digit is 4, number of possibilities = 3 X 3 X 2 = 18

  If last digit is 8, number of possibilities = 2 X 3 X 2 = 12

  Total = 30 ways.

   

  A simple way to prove that the answer is wrong:

  Total ways with no restriction except non-repetition = 5 X 4 X 3 X 2 = 120

  Total numbers less than 5000 = 2 X 4 X 3 X 2 = 48

  Numbers more than 5000 ( Both odd and even ) = 120 - 48 = 72, which is a contradiction to the answer given.

• wee_ws

  19 Feb 10, 16:04

  Hi,

    This is a nice question.

    We consider 2 cases.

    Case 1: 5-digit numbers (thanks to JDGao!)
    Number of possibilities = 4! x 2 = 48.
    Reason: 2 ways for last digit and 4! ways for 1st 4 digits.

    Case 2: 4-digit numbers
    There are 2 sub-cases to consider, which have been highlighted by Forbiddensinner already, i.e. 30 possibilities.

    Hence, we have a total of 78 such digits. Thanks.

  Cheers,
  Wen Shih

• Forbiddensinner

  19 Feb 10, 17:32

  Originally posted by wee_ws:

  Hi,

    This is a nice question.

    We consider 2 cases.

    Case 1: 5-digit numbers (thanks to JDGao!)
    Number of possibilities = 4! x 2 = 48.
    Reason: 2 ways for last digit and 4! ways for 1st 4 digits.

    Case 2: 4-digit numbers
    There are 2 sub-cases to consider, which have been highlighted by Forbiddensinner already, i.e. 30 possibilities.

    Hence, we have a total of 78 such digits. Thanks.

  Cheers,
  Wen Shih

  Alamak, I forgot to consider the 5 digit numbers!! 

  Looks like something which might come out for finite maths, I better be careful. 

  Thank you, Sir, for pointing it out.