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Questions regarding permutations .

  1. Dec 23, 2012 #1
    Questions regarding "permutations".

    1. The problem statement, all variables and given/known data

    These two mini questions , I think , can be adjusted in one thread itself.

    1. How many total 5-digit numbers divisible by 6 can be formed using 0,1,2,3,4,5 if repetition of digits is not allowed ?

    2. In the decimal system of numeration , find the number of 6-digit numbers in which the digit in any place is greater than the digit to the left of it.

    2. Relevant equations

    Concepts of permutations....

    3. The attempt at a solution

    Ok , so for the first one ,

    I made the cases as follows...

    For the divisibility by 6, the number should be divisible by 3 as well as 2.

    Case 1: Fix "0" at the end. The the sum can be 1+2+3+4+5 = 15
    I have to fill like this : __ __ __ __ 0
    Here 4 places have to be filled such that the sum yields the multiple of 3. If I subtract a number say x from 15 , I must get a multiple of 3 which is possible when x=3.
    Thus Permutations = 4! = 24

    Case 2 : Fix "2" at the end.
    __ __ __ __ 2
    The sum can be : 1+0+3+4+5 = 13
    I must have , 13-x = a multiple of 3
    So , x=1 , or x= 4

    Thus , permutations : 3x3x2x1 = 18
    As there are two sub cases , so Required permutations :18x2 = 36

    Case 3 : Fix "4".
    __ __ __ __ 4
    The sum can be : 1+2+3+0+5 = 11
    I must have , 11-x = a multiple of 3
    So , x=2 , x=5

    Thus here also Permutations = 36 (as there are two sub cases)

    Thus total Permutations : 36+36+24 = 96 , but answer given is 108 ! How ?

    For second one , I attempted for over 3 pages but failed !! I don't know how to do this !

    Please help !!

    Thanks in advance...:smile:
     
  2. jcsd
  3. Dec 23, 2012 #2

    HallsofIvy

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    Re: Questions regarding "permutations".

    Yes, and that means that the digits must sum to a multiple of 3 and the last digit must be even- 0, 2, or 4.

    [/quote]Case 1: Fix "0" at the end. The the sum can be 1+2+3+4+5 = 15
    I have to fill like this : __ __ __ __ 0
    Here 4 places have to be filled such that the sum yields the multiple of 3. If I subtract a number say x from 15 , I must get a multiple of 3 which is possible when x=3.[/quote]
    Okay, so the first four digits must be 1, 2, 4, 5, in any order.

    Right.

     
  4. Dec 23, 2012 #3
    Re: Questions regarding "permutations".

    Ok , but I have tried the so called "brute strength" , and have already filled 3-4 pages of my copy. It becomes too long.

    Can you suggest me any other way ? We get only 2-3 minutes of time in exam to solve such questions.
     
  5. Dec 23, 2012 #4

    I like Serena

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    Re: Questions regarding "permutations".

    Hi sankalp! :smile:

    I agree with HoI, I believe 96 is the correct answer.

    Suppose you pick 6 different digits and sort them.
    How many possibilities?
    And how many of those start with a zero? Because those won't count.
     
  6. Dec 23, 2012 #5

    ehild

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    Re: Questions regarding "permutations".

    You need to make five-digit numbers, so one of the six is not used. The number is divisible by three, so the sum of its digits must be divisible by 3. 0+1+2+3+4+5 = 15 is divisible, you can only left out 0 and 3.

    If you choose 1,2,3,4,5, 2 or 4 can be on the last place. How many possibilities you have to fill the other four places?

    If you choose 0,1,2,4,5, 0 the last digit should be 0, 2 or 4. Three cases to investigate.


    ehild
     
  7. Dec 23, 2012 #6
    Re: Questions regarding "permutations".

    Hii ILS !! :smile:

    (It has been a long time right ?)

    Alright may be there is a misprint in the answer given in my booklet for the first question , I asked here.

    10C6 right ? Because we have sorted all the permutations.

    So if I leave 0 , the combinations become 9C6.
    And yes 9C6 is the required answer !! :smile:

    Thanks ILS !!

    @ehild
    Hii ehild !! :smile:
    I already did the first question , and it has been point out correct by Halls of Ivy and ILS.
     
  8. Dec 23, 2012 #7

    haruspex

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    Re: Questions regarding "permutations".

    Except, Halls of Ivy and ILS both agreed with your 96. Following ehild's analysis (with which I agree) 108 is right: 48 leaving out 0, plus 72 leaving out 3 and allowing a leading zero, minus 12 with leading zero.
     
  9. Dec 23, 2012 #8

    ehild

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    Re: Questions regarding "permutations".

    Yes, 108 is the correct result. Leaving out zero, and putting 2 or 4 as the last digit, you get 4! possibilities for each, that is 2*24 = 48.

    Leaving out 3 the last digit can be 0 and the permutations of 4 for the other digits. Or the last digit can be either 2 or 4. Then you can choose 3 numbers as the first digit (0 is excluded) and 3! possibilities for the other digits: 4!+2*3*3! = 60

    ehild
     
  10. Dec 24, 2012 #9
    Re: Questions regarding "permutations".

    Oh ! I made a mistake.. When I made case 1 and case 2 , I forgot to add the digit which I fixed at the end. Thanks for your efforts !!

    :smile:
     
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