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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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theEric
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.mathsisfun.com/combinatorics/combinationspermutations.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!

alexwee123
 4 years ago
Best ResponseYou've already chosen the best response.0i think you would use combination

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i believe this is a permutation problem since it says "assemble" so \[nPr = \frac{n!}{(nr)!}\] so \[14P9 = \frac{14!}{(149)!}\] if im wrong then im a monkey s uncle

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that was supposed to be monkey's uncle...accidentally pressed enter

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0Try 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whats with the exclamation mark?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0exclamation mark means factorial

alexwee123
 4 years ago
Best ResponseYou've already chosen the best response.0say 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

alexwee123
 4 years ago
Best ResponseYou've already chosen the best response.0because you would still pick out the same magazines ..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[n! \implies n \times (n1) \times (n2) \times \cdots \times 1\] 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.

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1I 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@alexwee123 doesn't "assemble" mean arranging?

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{n!}{r!(nr)!}\] ?

alexwee123
 4 years ago
Best ResponseYou've already chosen the best response.0umm i don't really think so :/ assemble usually means to bring together, congregate

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm still not getting it. and ive been working on this problem for like 30mins ..

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1You'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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0assemble really feels like an arrangement...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i hate these fancy words =_=

alexwee123
 4 years ago
Best ResponseYou've already chosen the best response.0@lgbasallote but don't take my word for it :/ i'm pretty bad w/ differentiating between permutations and combinations

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1Agreed. 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...

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1If you agree, then we can help buddhababy002 with the problem if buddhababy002 still needs a hand! Haha!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me check my text book

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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)

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1You 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. \[nPr=\frac{n!}{(nr)!}\] where the "!" is actually a mathematical symbol that operates on the quantity before it like in the following\[8!=8*7*6*5*4*3*2*1\]\[5!=5*4*3*2*1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how would i put that into a calculator?

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1I 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i do have that key

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1Okay, now I'm not sure about your calculator. What happens when you type in the number "14" and hit the "!" button?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i get fact (14) 87178291200

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1That's factorial! So you got it to work! Will your calculator let you put in all of the formula? \[\frac{n!}{(nr)!}\] Where n is the total number of articles and r is the number of articles you want to use?

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1And do you want me to explain the formula, just for your own conceptual understanding?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me try the problem

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1I used an online site (wolframalpha.com) and got 726,485,760, so let me do it by hand quick..

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1I got the same answer. Maybe using the factorial key is complicated? Try writing down 14!, then (149)! = 5!, and then type in the division: of the 14! / 5!.

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1Hmm.. I see a 120! Haha!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got it.. lols thanks its so complicating

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1If 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?

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.mathsisfun.com/combinatorics/combinationspermutations.html seems to be helpful in understanding combinations and permutations.

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1I can also attempt to explain, if you'd like!

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1Explain the formula, for example.. But I don't know if I'll be able to explain it that well right now. I could try!

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1Maybe I could explain it well now!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think i kinda of got it.. thanks

theEric
 4 years ago
Best ResponseYou've already chosen the best response.1Alright! Good luck with everything! Thank you! :)
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