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Permutation and combination problem

  1. Nov 8, 2006 #1
    i recently had my A levels exam and was stuck at a question

    there are 8 balls in a box, 3 are similar and the rest are different, how many ways can 3 balls be chosen if the order of picking out the balls is not important

    its worth 4 marks, and i do not know where to start even.:frown:
     
  2. jcsd
  3. Nov 8, 2006 #2
    Consider the following: in how many possibilities do the similar balls appear?

    Say only ball 1 appears in spot 1 (ball 2 and 3 do not appear), you have (8-3)*(8-4) = 20 possibilities of the sort. Now say ball 1 appear in spot 2, you still have 20 possibitities. Now it's easy to see why the total number of possibilties is 3*20 = 60. Now consider ball 2 and 3. You have for each 60 possibities. However since ball 1, 2 and 3 are similar, all the possibilities that occur for ball 2 and 3 have equivalents in the possibilities for ball 1, thus we only take 60 into account.

    Next, use the same procedure for 2 balls appearing:In how many ways can 2 balls be rearanged with 3 spots? The awnser is 3*2 = 6. For the remaning spot, how many possibilities can you have? The awnser is 8-3 = 5. Now with using combinations, the number of possibilities is 6*5 = 30. Now since we are considering 2 balls, we are dealing with reoccuring possibilities (i.e. ball 1 inversed with ball 2 etc.). The question is, if there's 2 balls, in how many ways can they be interchanged? The awnser is simply 2*1 = 2. Since we do not wish to have reoccurences, we devided 30 by 2, giving 15.

    We then consider the 3 similar balls showing up. It should be obvious that this account for only 1 possibility.

    Now we have 60 + 15 + 1 = 76 as the final awnser.
     
  4. Nov 9, 2006 #3
    thanks

    thanks for the reply, but i thought the order did not matter? why would it be 20*3 and not just 20 since order does not matter.
     
  5. Nov 9, 2006 #4
    Hummm... yeah true, since it says the order dosen't count, it's 20. For 2 balls, since the order dosen't count it's only 5... so the final awnser should be 26.
     
  6. Nov 10, 2006 #5
    thanks

    thats what i thought too......but...... if you look closer at the question, it says the order the ball was picked does not matter, that does not mean the order it will be after all 3 were picked did not matter, am i right to say that?
     
  7. Nov 10, 2006 #6

    acm

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    Ways of choosing 3 balls from eight : (8,3)
    Ways of choosing 3 balls from eight when three are the same : (8,3)/3!
    8!/5!(3!)^2 =336/36

    Probably not right, just posting my thoughts.
     
  8. Nov 12, 2006 #7
    How can you have non-integer number of possibilities? :tongue2:
     
  9. Nov 12, 2006 #8
    I think 26 is the correct anwnser... If they specify that the order dosen't count, it means that different arrangements for the same balls are neglected.
     
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