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buddhababy002

Can someone help me with this problem? A mathematics journal has accepted 14 articles for publication. However, due to budgetary restraints only 9 articles can be published this month. How many ways can the journal editor assemble 9 of the 14 articles for publication?

  • one year ago
  • one year ago

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  1. theEric
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    http://www.mathsisfun.com/combinatorics/combinations-permutations.html Has some information on combinations and permutations. I think I know the answer but I have to double check in case I'm wrong!

    • one year ago
  2. alexwee123
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    i think you would use combination

    • one year ago
  3. lgbasallote
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    i believe this is a permutation problem since it says "assemble" so \[nPr = \frac{n!}{(n-r)!}\] so \[14P9 = \frac{14!}{(14-9)!}\] if im wrong then im a monkey s uncle

    • one year ago
  4. lgbasallote
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    that was supposed to be monkey's uncle...accidentally pressed enter

    • one year ago
  5. Kainui
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    Try scaling the problem down to figure out a method of how to do it to understand how with a fewer so you can draw out the different ways with circles representing the articles. Playing around mathematics in this way will give you the better understanding of why you're doing all this factorial stuff.

    • one year ago
  6. buddhababy002
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    whats with the exclamation mark?

    • one year ago
  7. lgbasallote
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    exclamation mark means factorial

    • one year ago
  8. alexwee123
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    say that there are magazines a,b,c,d and say you need to pick out 3 of them but it really doesn't matter whether you pick out magazines a,b,c or b,a,c, order doesn't really matter

    • one year ago
  9. alexwee123
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    because you would still pick out the same magazines ..

    • one year ago
  10. lgbasallote
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    \[n! \implies n \times (n-1) \times (n-2) \times \cdots \times 1\] for example \[5! \implies 5 \times 4 \times 3 \times 2 \times 1\] \[4! \implies 4\times 3 \times 2 \times 1\] \[8! \implies 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\] etc.

    • one year ago
  11. theEric
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    I think it's a combination, since you don't want to pick magazines A, B, and C and then also consider the possibility B, C, and A!

    • one year ago
  12. alexwee123
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    14C9 @lgbasallote

    • one year ago
  13. lgbasallote
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    @alexwee123 doesn't "assemble" mean arranging?

    • one year ago
  14. theEric
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    \[\frac{n!}{r!(n-r)!}\] ?

    • one year ago
  15. alexwee123
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    umm i don't really think so :/ assemble usually means to bring together, congregate

    • one year ago
  16. buddhababy002
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    I'm still not getting it. and ive been working on this problem for like 30mins ..

    • one year ago
  17. theEric
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    You're right, I suppose so! I don't do well with math speak. I assumed that meant "choose", but assemble would probably mean permutation, and I have a feeling you'd know better than I.

    • one year ago
  18. lgbasallote
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    assemble really feels like an arrangement...

    • one year ago
  19. lgbasallote
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    i hate these fancy words =_=

    • one year ago
  20. alexwee123
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    @lgbasallote but don't take my word for it :/ i'm pretty bad w/ differentiating between permutations and combinations

    • one year ago
  21. theEric
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    Agreed. I looked up an example that specifically mentions how the problem is using "assemble", and it agrees with lgbasallote. "Determine how many ways you can assemble a pizza with ONLY three toppings (pepperoni, sausage, bacon). This will depend on the order that ingredients are placed on the pizza. For example, putting on pepperoni, then sausage, then bacon is different than putting on bacon, then pepperoni, then sausage." So I assume that order matters, and his equation is appropriate. This... is math...

    • one year ago
  22. theEric
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    If you agree, then we can help buddhababy002 with the problem if buddhababy002 still needs a hand! Haha!

    • one year ago
  23. buddhababy002
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    let me check my text book

    • one year ago
  24. buddhababy002
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    Well, in my text book they give me two equations to use the first one is \[n ^{P}r\] (Permutation notation) and this one \[n^{C}r\] (Combination Notation)

    • one year ago
  25. theEric
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    You will want to use the formula lgbasallote gave if it is a situation where you want to consider different combinations of the same 9 articles. I'll post it here so it's lower and we don't have to scroll so much to see it. \[nPr=\frac{n!}{(n-r)!}\] where the "!" is actually a mathematical symbol that operates on the quantity before it like in the following\[8!=8*7*6*5*4*3*2*1\]\[5!=5*4*3*2*1\]

    • one year ago
  26. buddhababy002
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    how would i put that into a calculator?

    • one year ago
  27. theEric
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    I believe we should use \[nPr\] for the same reason lgbasallote stated; the problem used the word "assemble", indicating we should count the same elements (articles) in different orders. Like, maybe the guy want to know how many ways he can stack his pile of 9 articles. Do you have a "!" key?

    • one year ago
  28. buddhababy002
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    okay.. let me check

    • one year ago
  29. buddhababy002
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    yes i do have that key

    • one year ago
  30. theEric
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    Okay, now I'm not sure about your calculator. What happens when you type in the number "14" and hit the "!" button?

    • one year ago
  31. buddhababy002
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    i get fact (14) 87178291200

    • one year ago
  32. theEric
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    That's factorial! So you got it to work! Will your calculator let you put in all of the formula? \[\frac{n!}{(n-r)!}\] Where n is the total number of articles and r is the number of articles you want to use?

    • one year ago
  33. theEric
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    And do you want me to explain the formula, just for your own conceptual understanding?

    • one year ago
  34. buddhababy002
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    let me try the problem

    • one year ago
  35. theEric
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    Okay.

    • one year ago
  36. buddhababy002
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    i got 120

    • one year ago
  37. theEric
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    I used an online site (wolframalpha.com) and got 726,485,760, so let me do it by hand quick..

    • one year ago
  38. theEric
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    Actually, calculator.

    • one year ago
  39. theEric
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    I got the same answer. Maybe using the factorial key is complicated? Try writing down 14!, then (14-9)! = 5!, and then type in the division: of the 14! / 5!.

    • one year ago
  40. theEric
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    87178291200 / 120

    • one year ago
  41. theEric
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    Hmm.. I see a 120! Haha!

    • one year ago
  42. buddhababy002
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    i got it.. lols thanks its so complicating

    • one year ago
  43. theEric
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    If you're using a calculator for a test, it's good to know how you have to deal with it, haha. No problem. Anything else I can help you with?

    • one year ago
  44. theEric
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    http://www.mathsisfun.com/combinatorics/combinations-permutations.html seems to be helpful in understanding combinations and permutations.

    • one year ago
  45. theEric
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    I can also attempt to explain, if you'd like!

    • one year ago
  46. theEric
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    Explain the formula, for example.. But I don't know if I'll be able to explain it that well right now. I could try!

    • one year ago
  47. theEric
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    Maybe I could explain it well now!

    • one year ago
  48. buddhababy002
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    i think i kinda of got it.. thanks

    • one year ago
  49. buddhababy002
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    :D

    • one year ago
  50. theEric
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    Alright! Good luck with everything! Thank you! :)

    • one year ago
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