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Easy combinations question

  1. Nov 22, 2013 #1

    joshmccraney

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    hey all

    if i have, say 10 y's, 5 x's, and 4 z's and i want to see how many groups of 6 i can make with at least 5 y's, i know the answer is [itex]{10 \choose 5}{9 \choose 1}+{10 \choose 6}[/itex] which is the combinations of 5 x's added to the combinations of 6 x's.

    but why isn't the answer [itex]{10 \choose 5}{14 \choose 1}[/itex], which is the 5 x's and then the remaining 9 non-x characters added to the last 5 x-characters?

    please help!

    thanks!

    josh
     
  2. jcsd
  3. Nov 23, 2013 #2

    mathman

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    Your statement has many typos making it hard to understand. x's and y's are mixed up.
     
  4. Nov 24, 2013 #3

    joshmccraney

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    sorry about this

    this should make more sense:

    if i have, say 10 y's, 5 x's, and 4 z's and i want to see how many groups of 6 i can make with at least 5 y's, i know the answer is [itex]{10 \choose 5}{9 \choose 1}+{10 \choose 6}[/itex] which is the combinations of 5 y's added to the combinations of 6 y's.

    but why isn't the answer [itex]{10 \choose 5}{14 \choose 1}[/itex], which is the 5 y's and then the remaining 9 non-y characters added to the last 5 y-characters?
     
  5. Nov 25, 2013 #4

    Office_Shredder

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    Your solution overcounts the ways that you could have picked 6 y's. Let's suppose the y's are numbered 1 through 10. Your solution says first if I pick five y's, say 1,2,3,4,5, then I can pick any one of the remaining 14 characters and get a distinct solution. For example I could pick y number 6.

    On the other hand, I could start with five y's, 2,3,4,5,6, and then out of the remaining 14 characters I could pick y number 1, and end up with the exact same set of y's as above. Your attempt counts these as two separate ways of picking 1,2,3,4,5,6 out of my set of y's.
     
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