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buddhababy002 Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. theEric Group Title
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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!

    • 2 years ago
  2. alexwee123 Group Title
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    i think you would use combination

    • 2 years ago
  3. lgbasallote Group Title
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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

    • 2 years ago
  4. lgbasallote Group Title
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    that was supposed to be monkey's uncle...accidentally pressed enter

    • 2 years ago
  5. Kainui Group Title
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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.

    • 2 years ago
  6. buddhababy002 Group Title
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    whats with the exclamation mark?

    • 2 years ago
  7. lgbasallote Group Title
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    exclamation mark means factorial

    • 2 years ago
  8. alexwee123 Group Title
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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

    • 2 years ago
  9. alexwee123 Group Title
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    because you would still pick out the same magazines ..

    • 2 years ago
  10. lgbasallote Group Title
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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.

    • 2 years ago
  11. theEric Group Title
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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!

    • 2 years ago
  12. alexwee123 Group Title
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    14C9 @lgbasallote

    • 2 years ago
  13. lgbasallote Group Title
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    @alexwee123 doesn't "assemble" mean arranging?

    • 2 years ago
  14. theEric Group Title
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    \[\frac{n!}{r!(n-r)!}\] ?

    • 2 years ago
  15. alexwee123 Group Title
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    umm i don't really think so :/ assemble usually means to bring together, congregate

    • 2 years ago
  16. buddhababy002 Group Title
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    I'm still not getting it. and ive been working on this problem for like 30mins ..

    • 2 years ago
  17. theEric Group Title
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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.

    • 2 years ago
  18. lgbasallote Group Title
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    assemble really feels like an arrangement...

    • 2 years ago
  19. lgbasallote Group Title
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    i hate these fancy words =_=

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

    • 2 years ago
  21. theEric Group Title
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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...

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

    • 2 years ago
  23. buddhababy002 Group Title
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    let me check my text book

    • 2 years ago
  24. buddhababy002 Group Title
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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)

    • 2 years ago
  25. theEric Group Title
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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\[8!=8*7*6*5*4*3*2*1\]\[5!=5*4*3*2*1\]

    • 2 years ago
  26. buddhababy002 Group Title
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    how would i put that into a calculator?

    • 2 years ago
  27. theEric Group Title
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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?

    • 2 years ago
  28. buddhababy002 Group Title
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    okay.. let me check

    • 2 years ago
  29. buddhababy002 Group Title
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    yes i do have that key

    • 2 years ago
  30. theEric Group Title
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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?

    • 2 years ago
  31. buddhababy002 Group Title
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    i get fact (14) 87178291200

    • 2 years ago
  32. theEric Group Title
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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?

    • 2 years ago
  33. theEric Group Title
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    And do you want me to explain the formula, just for your own conceptual understanding?

    • 2 years ago
  34. buddhababy002 Group Title
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    let me try the problem

    • 2 years ago
  35. theEric Group Title
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    Okay.

    • 2 years ago
  36. buddhababy002 Group Title
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    i got 120

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

    • 2 years ago
  38. theEric Group Title
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    Actually, calculator.

    • 2 years ago
  39. theEric Group Title
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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!.

    • 2 years ago
  40. theEric Group Title
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    87178291200 / 120

    • 2 years ago
  41. theEric Group Title
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    Hmm.. I see a 120! Haha!

    • 2 years ago
  42. buddhababy002 Group Title
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    i got it.. lols thanks its so complicating

    • 2 years ago
  43. theEric Group Title
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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?

    • 2 years ago
  44. theEric Group Title
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    http://www.mathsisfun.com/combinatorics/combinations-permutations.html seems to be helpful in understanding combinations and permutations.

    • 2 years ago
  45. theEric Group Title
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    I can also attempt to explain, if you'd like!

    • 2 years ago
  46. theEric Group Title
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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!

    • 2 years ago
  47. theEric Group Title
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    Maybe I could explain it well now!

    • 2 years ago
  48. buddhababy002 Group Title
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    i think i kinda of got it.. thanks

    • 2 years ago
  49. buddhababy002 Group Title
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    :D

    • 2 years ago
  50. theEric Group Title
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    Alright! Good luck with everything! Thank you! :)

    • 2 years ago
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