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anonymous

  • one year ago

MEDAL+FAN+TESTIMONIAL The calculation 9!/3!6! is equivalent to 9C3 (or 9C6) Explain clearly why you could solve this question using combinations, and why this is equivalent to considering permutations with repeated items.

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. anonymous
    • one year ago
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    @nincompoop

  3. anonymous
    • one year ago
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    Oh thanks!

  4. anonymous
    • one year ago
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    Thanks for your previous explanations! They were easy to understand:)

  5. anonymous
    • one year ago
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    What can I do to get an overview of this question?

  6. ganeshie8
    • one year ago
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    Hey is that the complete problem ?

  7. anonymous
    • one year ago
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    We have looked at situations in which we need to determine the number of possible routes between two places. We can look at the situation below as 9 steps, six of which must be East and three of which must be South. This gives us 9!/3!6! = 84 possible routes

  8. anonymous
    • one year ago
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    1 Attachment
  9. anonymous
    • one year ago
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    Those are the starting places:)

  10. anonymous
    • one year ago
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    Thanks for pointing them out

  11. anonymous
    • one year ago
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    The file "imagen" is exactly as it is

  12. ganeshie8
    • one year ago
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    |dw:1437035983074:dw|

  13. anonymous
    • one year ago
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    What may be the discernible connection here?

  14. ganeshie8
    • one year ago
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    Lets work this problem using "permutations" only first

  15. anonymous
    • one year ago
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    Ok

  16. UsukiDoll
    • one year ago
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    maybe we need to calculate the 9!/3!6! first?

  17. anonymous
    • one year ago
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    6 east and 3 south.

  18. ganeshie8
    • one year ago
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    A valid route is 9 steps long. Must have 6 East and 3 South, yes ?

  19. anonymous
    • one year ago
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    Yes. They have to be the shortest.

  20. UsukiDoll
    • one year ago
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    I got 6 east 3 south too

  21. ganeshie8
    • one year ago
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    Like : \[EEEEEE~SSS\]

  22. anonymous
    • one year ago
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    9P9?

  23. ganeshie8
    • one year ago
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    How many distinct words can you make by rearranging the characters in that string above ?

  24. anonymous
    • one year ago
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    9!/(6!*3!)

  25. anonymous
    • one year ago
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    Because of the indistinguishable letters

  26. ganeshie8
    • one year ago
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    How did you get that, could you please explain a bit more..

  27. UsukiDoll
    • one year ago
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    wasn't that 9!/(6!)(3!) from the original problem?

  28. anonymous
    • one year ago
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    Because there are 9 letters to be arranged, but permutation 9! would have to be reduced by letters that are identical to each other. And 6! accounts for Es and 3! accounts for Ss. Therefore the total number of distinct words are divided by overlapping ones.

  29. anonymous
    • one year ago
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    I had a similar question before where I had to rearrange letters to make distinct words, and one of the math tutors here showed me that I need to divide the original all possible options by the factorials of overlapping ones.

  30. ganeshie8
    • one year ago
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    Perfect!

  31. ganeshie8
    • one year ago
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    so that is a method using permutations only, lets look at working the total number of routes using "combinations" only

  32. anonymous
    • one year ago
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    For example, in "mathematics" there are 2 "m"s 2 identical "a"s 2 identical "t"s So 11!/(2!*2!*2!)=possible rearranged distinct words from mathematics.

  33. anonymous
    • one year ago
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    Ok.

  34. UsukiDoll
    • one year ago
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    \[\frac{n!}{r!(n!-r!)}\] is the formula for the number of possible combinations of r objects in from the set of n objects it's usually written in nCr.

  35. ganeshie8
    • one year ago
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    To go from red point on top left corner to the green point on bottom right corner, you need to take 9 steps. 3 of them South and 6 toward East. So if you `choose 3 steps for South out of total 9 steps`, the remaining 6 steps are forced to be East. |dw:1437036960970:dw|

  36. anonymous
    • one year ago
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    3 steps south where it could be anywhere, but 6 steps east would have to be made to get to the destination

  37. ganeshie8
    • one year ago
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    |dw:1437037029139:dw|

  38. UsukiDoll
    • one year ago
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    so if we have 9C3 that would mean that we let n = 9 and r=3 for nCr \[\frac{9!}{3!(9!-3!)}\]

  39. anonymous
    • one year ago
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    How come it's the 3! down below 9!?

  40. anonymous
    • one year ago
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    I know that 3 souths and 6 easts. But somehow 6 is ignored here?

  41. anonymous
    • one year ago
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    I understand Ganeshie's easy explanations, but suddenly formatting them in an equation is quite mind numbing.

  42. ganeshie8
    • one year ago
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    |dw:1437037261640:dw|

  43. anonymous
    • one year ago
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    3 steps must be south somewhere, and 6 steps must be east. Ok.

  44. anonymous
    • one year ago
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    The order in which they are taken does not matter?

  45. anonymous
    • one year ago
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    So long as the 3 steps to the south and 6 steps to the east are achieved .

  46. ganeshie8
    • one year ago
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    Let me ask you a related question, Pretend that you have 9 girls as friends, you want to choose 1 girl to go out with. How many possibilities are there to pick one girl out of 9 girls ?

  47. anonymous
    • one year ago
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    9 girls.

  48. anonymous
    • one year ago
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    because there are 9 girls, from whom to pick one.

  49. ganeshie8
    • one year ago
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    Right, that means you're rejecting 8 other girls right ?

  50. anonymous
    • one year ago
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    Yes.

  51. anonymous
    • one year ago
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    So if I pick a certain path I would be rejecting others.

  52. ganeshie8
    • one year ago
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    As you can see, this question is not that retarded. This is actually a combination problem. Number of ways of choosing 1 girl from 9 girls = 9C1 = 9 Here we are not worrying about 8 other girls. They are forced to be rejected automatically.

  53. UsukiDoll
    • one year ago
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    WOW ganeshie!!! that r word!

  54. ganeshie8
    • one year ago
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    Haha! @Robert136 just want you see that when you choose 3 steps for South, the remaining 6 other steps are automatically forced to be rejected for South.

  55. UsukiDoll
    • one year ago
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    true you can only go one way after you go 3 steps south |dw:1437037748310:dw|

  56. UsukiDoll
    • one year ago
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    |dw:1437037774091:dw|

  57. ganeshie8
    • one year ago
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    So the math is simple : Total number of routes = number of ways of choosing 3 steps for South out of 9 total steps

  58. anonymous
    • one year ago
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    and I put focus on 3 steps for south because moving east is a predetermined act?

  59. ganeshie8
    • one year ago
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    In the exact same spirits, Total number of routes = number of ways of choosing 6 steps for East out of 9 total steps

  60. ganeshie8
    • one year ago
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    Yes, when you choose 3 steps for south, the fate of remaining 6 steps is forced.

  61. anonymous
    • one year ago
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    9C3

  62. UsukiDoll
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @UsukiDoll so if we have 9C3 that would mean that we let n = 9 and r=3 for nCr \[\frac{9!}{3!(9!-3!)}\] \(\color{blue}{\text{End of Quote}}\)

  63. ganeshie8
    • one year ago
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    Yes, that also equals 9C6. With the earlier analogy, "selecting 1 girl out of 9 girls" is same as "rejecting the 8 other girls".

  64. UsukiDoll
    • one year ago
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    \[\frac{9!}{3!(9!-3!)} \rightarrow \frac{9!}{3!(6!)}\]

  65. anonymous
    • one year ago
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    So the reason why both permutation and combination could both be used is because the circumstances in which the question is asked permits the use of both.

  66. anonymous
    • one year ago
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    Or that conceptually both combination and permutation are relevant in deriving the answer.

  67. anonymous
    • one year ago
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    Do I still appear retarded? Are my conclusions way offXD?

  68. UsukiDoll
    • one year ago
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    similarly suppose we do have 9C6 then n=9 and r =6 \[\frac{9!}{6!(9!-6!)} \rightarrow \frac{9!}{6!(3!)}\] same thing but different order

  69. anonymous
    • one year ago
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    I see. But that is a coincidence no? The mysteries of numbers

  70. UsukiDoll
    • one year ago
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    tbh I haven't went into much depth on it.. permutations <-order matters! combinations <- order doesn't matter

  71. anonymous
    • one year ago
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    Like anything with the terms of combination method that is arrived at the same factorials

  72. UsukiDoll
    • one year ago
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    formula for permutations is similar to the combination except the r! is missing

  73. anonymous
    • one year ago
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    Usuki you still have me blocked I can't write you testimonials yet

  74. anonymous
    • one year ago
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    XD

  75. UsukiDoll
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @UsukiDoll \[\frac{n!}{r!(n!-r!)}\] is the formula for the number of possible combinations of r objects in from the set of n objects it's usually written in nCr. \(\color{#0cbb34}{\text{End of Quote}}\) For permutations \[\frac{n!}{n!-r!}\] or \[\frac{n!}{(n-r)!} \] written in the form nPr

  76. UsukiDoll
    • one year ago
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    actually that sounds related.. ummm you can refer to this link @Pci https://www.mathsisfun.com/combinatorics/combinations-permutations.html

  77. anonymous
    • one year ago
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    Gracias a todos por comentarios muy entiendibles.

  78. anonymous
    • one year ago
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    If anyone needs Spanish practice in return:)

  79. ganeshie8
    • one year ago
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    Lets go back to the earlier string : \[EEEEEE~SSS\] Consider the problem of finding number of total 9 letter distinct words by rearranging letters in above string. How many ways can you choose 3 places for S's ? |dw:1437038371595:dw|