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Combinatorics problem, apples and pears

  1. Feb 8, 2017 #1
    1. The problem statement, all variables and given/known data
    There are 15 different apples and 10 different pears. How many ways are there for Jack to pick an apple or a pear and then for Jill to pick an apple and a pear?

    2. Relevant equations


    3. The attempt at a solution
    Let apples be A and pears P. Since Jack chooses either an apple or pear, he has 25 possibilities. This leaves Jill to choose a pair from either:
    14A, 10P or
    15A, 9P

    So Jill's choices = 25[14(10) + 15(9)] = 6875

    Which is rather far from the given answer of 15[10(14 + 9)] = 3450 and I do not see how this was arrived at. What am I doing wrong here?
     
  2. jcsd
  3. Feb 8, 2017 #2

    PeroK

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    Does it make a difference what Jack chooses. You have assumed not.

    Does it make a difference who chooses first? If not, could you double check your answer by letting Jill choose first?
     
  4. Feb 8, 2017 #3
    If Jill chose first the answer would be 150. The problem does specify that jack goes first so we can discount this option.
    As for whether Jack's choice makes a difference, I thought the 14A, 10P and 15A, 9P split dealt with that?
     
  5. Feb 8, 2017 #4

    PeroK

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    If Jack chooses, say, the first apple, then you have calculated 140 + 135 options for Jill, which is almost twice the options she has if she chose first. And the same for all Jack's choices.

    In particular the 140 includes apple 2 and pear 1 and the 135 included this option again, but also choices with apple 1, which shouldn't be possible.

    note that it makes no difference who chooses first. Think about it.
     
  6. Feb 8, 2017 #5
    So in fact Jill only has 150 distinct pairs available, regardless of Jack's selection?
     
  7. Feb 8, 2017 #6

    PeroK

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    At most 150. And always less than 150 if Jack chooses first.
     
  8. Feb 8, 2017 #7
    Yes, I can see that with the help of a quick sketch that the given answer counts the same pairs multiple times and is thus incorrect. Thanks for the clarity.
     
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