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How many ways can you choose 3 cards so the total is 15?

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data

    The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are printed on individual cards.

    How many ways can you choose three cards so that the total of the three numbers is 15?

    3. The attempt at a solution

    The different ways I can see are:

    0, 9, 6
    0, 8, 7
    1, 9, 5
    1, 8, 6
    1, 5, 9
    2, 9, 4
    2, 8, 5
    2, 7, 6
    2, 6, 7
    2, 4, 9
    3, 8, 4
    3, 7, 5
    4, 9, 2
    4, 8, 3
    4, 6, 5

    Is 15 different ways the answer?
     
  2. jcsd
  3. May 18, 2012 #2
    Is 15 the right answer though?
     
  4. May 18, 2012 #3
    Does the order of picking the cards matter?

    Also, is there replacement?


    The question seems a little ambiguous.
     
  5. May 18, 2012 #4
    Maybe you put it this way.
    0+(15)
    1+(14)
    2+(13)
    3+(12)
    4+(11)
    5+(10)
    ..
    ..
    9+(6)
    where in bracket is the sum of 2 numbers.
     
  6. May 19, 2012 #5
    Another, generalized way to solve this problem would be to use Multinomial Theorem. The answer you need would be the coefficient of x15 in

    [itex](1+x+x^2+......+x^9)(1+x+x^2+......+x^9)(1+x+x^2+......+x^9)[/itex]

    which is equal to

    [itex](1+x+x^2+......+x^9)^{3}[/itex]

    This can be applied to any similar problem with any kind of conditions on the numbers, and its easier than listing all the numbers out, especially when numbers get bigger :wink:

    If there is no replacement, you need to subtract the cases where the power of x is same.
     
  7. May 19, 2012 #6
    It does not mention about replacement in the question.

    I think the answer is 15. Could someone please just say yes or no? As I have attempted it...
     
  8. May 19, 2012 #7
    The answer is wrong. :smile:


    Edit : Also, as Villyer asked, does the order of picking the cards matter? Your answer is wrong in either case, though...
     
    Last edited: May 19, 2012
  9. May 19, 2012 #8
    It is not mentioned in the question at all. My gut feeling is that it does matter...
     
  10. May 19, 2012 #9
    Okay, try solving for both cases.

    Case 1 : It matters.
    Case 2 : It doesn't.

    Lets say your initial solution was for Case 1.

    Why your solution is wrong, is because you have
    Now there needs to be a few more ways that consider 5,1,9 and 5,9,1 and 9,1,5 and 9,5,1. This can be done to each of the numbers you listed down, so that would result in lots more cases than you already have.


    If you had solved it for Case 2, then the 1,9,5 and 1,5,9 would mean the same. So you would have to eliminate such numbers from your answer.


    PS : the number of Case 1 has a special relation with Case 2, so just doing Case 2 will suffice, after which you can multiply the 'extra ways.'
     
  11. May 19, 2012 #10
    Ok so I am only going to look at the case 2, because I don't think it matters actually. I did say it did sorry in an earlier post, I actually meant it didn't.

    So surely these are the only 15 ways if not, could someone at least give me one more?...

    0, 9, 6
    0, 8, 7
    1, 9, 5
    1, 8, 6
    1, 5, 9
    2, 9, 4
    2, 8, 5
    2, 7, 6
    2, 6, 7
    2, 4, 9
    3, 8, 4
    3, 7, 5
    4, 9, 2
    4, 8, 3
    4, 6, 5
     
  12. May 19, 2012 #11
    Did you read the latter part of my previous post?? :confused:

     
  13. May 19, 2012 #12
    so for case 1, the answer is 15 and case 2 it is 13 or is it case 1 = 13 and case 2 = 15
     
  14. May 19, 2012 #13
    Why 5,1,9 not counted?
     
  15. May 19, 2012 #14
    Is Case 2 correct like this with 10 ways? Or have I forgotten any?

    0, 9, 6
    0, 8, 7
    1, 9, 5
    1, 8, 6
    2, 9, 4
    2, 8, 5
    2, 7, 6
    3, 8, 4
    3, 7, 5
    4, 6, 5

    God knows how many there are for Case 1, I won't put an answer in the box, never mind.
     
  16. May 19, 2012 #15
    That looks fine, although you probably could have made it easier (on yourself) to check by presenting each three-card set in ascending order.

    As for the case where the order of selection matters, you just have to consider how many arrangements of each of the selections you have here can be chosen in.
     
  17. May 19, 2012 #16
    0, 6, 9
    0, 7, 8
    1, 5, 9
    1, 6, 8
    2, 4, 9
    2, 5, 8
    2, 6, 7
    3, 4, 8
    3, 5, 7
    4, 5, 6

    so that's 10 ways for Case 2.

    For Case 1 is it just a matter of 10 by 6 as each line has 6 permutations possible? Hence, Case 1 = 60 different ways
     
  18. May 19, 2012 #17
    That's it. And also I think that no replacement is implied by the question, so you're done.
     
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