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Permutations homework question

  1. Nov 5, 2005 #1
    i just want to ask if my answer is correct...
    the problems is .... how many numbers greater than 300 000 are there using only the digits 1,1,1,2,2,3?

    my answer is... 1(three)*5(the position)*2(the numbers 1 and 2) = 10
    therefore, the numbers greater than 300 000 is 10

    is this right? thanks for the replies...
     
    Last edited: Nov 5, 2005
  2. jcsd
  3. Nov 5, 2005 #2

    NateTG

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    Hmm...
    111223
    111232
    111322
    113122
    131122
    311122
    311221
    311212
    312112
    321112
    321121

    Clearly there are more than 10.
     
  4. Nov 5, 2005 #3
    i think you are misunderstanding the question... the ways to get combinations greater than 300,000

    this ones are right... but im asking if it is ten?
     
  5. Nov 5, 2005 #4
    its 1 * 5! , I think at least. the 3 has to be in front, and then you got 5 more numbers that can go anywhere. so 5! is 120. Therefore 120 numbers greater than 300,000.
     
  6. Nov 5, 2005 #5
    i dont think so cause you have 3 ones and 2 twos... i know what you mean, but the number is not different and that is applicable only if the numbers is different
     
    Last edited: Nov 5, 2005
  7. Nov 5, 2005 #6
    Oh ok, my bad. Let me think:D
     
  8. Nov 5, 2005 #7
    5!/(3! x 2!)

    is the answer im pretty sure. which is 10 which means you're right.
     
  9. Nov 5, 2005 #8
    for the first time, lol, im right, *claps*
     
  10. Nov 5, 2005 #9

    HallsofIvy

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    There are, of course, an infinite number of numbers "larger than 300000" with any digits at all!
    You seem to be assuming "larger than 300000 but less than 400000". If that was the problem, then it would help a lot for you to say so!
     
  11. Nov 5, 2005 #10
    ok then thanks for the info
     
  12. Nov 6, 2005 #11

    HallsofIvy

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    Now that I read it again, I see that I misunderstoond the question. I was thinking you meant 'numbers larger than 300000 using the digits "1", "2", and "3"' but since you are repeating the "1", you clearly mean just using those 6 digits.
    To be larger than 300000, the first digit must be 3 so it's really "in how many ways can we write the 5 digits 1, 1, 1, 2, 2". That's really a "binomial" problem: it is [itex]\frac{5!}{3!2!}= 10[/itex]. One way of seeing that is to imagine that each of those had a subscript: 11, 12, 13, 21, 22 so that they are distinguishable. There are 5! ways of ordering those 5 symbols. Now we not that 3! ways of rearranging just the 3 "1"s so that for each of the ways of rearranging those 5 "distinguishable" symbols there are 5 more that are just rearranging the different "1"s. Since they are NOT distinguishable, we don't want to count those as 6 different ways so we divide by 3!= 6. The same thing happens with the "2"s. In order NOT to count as different orders that just swap the two "2"s, we divide by 2!= 2.
     
  13. Nov 6, 2005 #12
    thanks again for the info... it help a lot on my understanding regarding permutations... can you check my other post, which is "probability".
     
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