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Doing some how many combinations calculation

  1. Mar 14, 2009 #1
    Doing some "how many combinations" calculation

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

    We're given the number 123215.

    How many different numbers can we get by changing the order of the digits?

    This is a six-digit number, so there are six positions for each digit.


    This is an unsorted selection with covering (not sure if that's what you call it in English).



    3. The attempt at a solution

    I've tried nCr, nPr and pretty much everything that's tryable;) I just can't get it right. I know for a fact that the correct answer is 180.
     
  2. jcsd
  3. Mar 14, 2009 #2
    Re: Doing some "how many combinations" calculation

    Do you know what nCr physically means? Well, it tells you how many ways you can select (without ordering) r different objects, from a set of n objects.

    This is not what you have to do here. you have Order the given 6 digits, in order to produce newer numbers.

    Order, and selection is done by nPr. Now you have to select and order 6 digits, from a set of 6 digits. So you must use 6P6.

    But wait!!! That only works if all the digits are different. Here there are two 2's and two 1's. Can you account for that?
     
  4. Mar 15, 2009 #3
    Re: Doing some "how many combinations" calculation

    So I'm going to order a set of 6 digits. There are 4 different digits, but two 2s and two 1s. 6 digits total. The order of the 2s and the 1s doesn't matter.

    My first guess was 6P4, but that can't be right.
     
  5. Mar 15, 2009 #4
    Re: Doing some "how many combinations" calculation

    Think it out logically. Let's look at the 1s only. Every number you create will contain these 2 digits whose order you can transpose without affecting the whole number. That means there are exactly two of these meaningless rogue permutations per whole number, so we can count properly by first counting all the permutations pretending each number is distinct and then dividing by 2. Can you see how to get rid of the rest of the rogues pertaining to the two 2s?
     
  6. Mar 15, 2009 #5
    Re: Doing some "how many combinations" calculation

    ahh, so we divide by 2 for the 1s, and then divide by 2 again for the 2s. That gives 180:D

    Not really logic to me:S Probability is a weird branch of maths.

    And thanks for the help! Appriciated it.
     
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