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

- chestercat

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- theEric

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!

- alexwee123

i think you would use combination

- lgbasallote

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

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## More answers

- lgbasallote

that was supposed to be monkey's uncle...accidentally pressed enter

- Kainui

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.

- anonymous

whats with the exclamation mark?

- lgbasallote

exclamation mark means factorial

- alexwee123

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

- alexwee123

because you would still pick out the same magazines ..

- lgbasallote

\[n! \implies n \times (n-1) \times (n-2) \times \cdots \times 1\]
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.

- theEric

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!

- alexwee123

14C9 @lgbasallote

- lgbasallote

@alexwee123 doesn't "assemble" mean arranging?

- theEric

\[\frac{n!}{r!(n-r)!}\]
?

- alexwee123

umm i don't really think so :/
assemble usually means to bring together, congregate

- anonymous

I'm still not getting it. and ive been working on this problem for like 30mins ..

- theEric

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.

- lgbasallote

assemble really feels like an arrangement...

- lgbasallote

i hate these fancy words =_=

- alexwee123

@lgbasallote but don't take my word for it :/
i'm pretty bad w/ differentiating between permutations and combinations

- theEric

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...

- theEric

If you agree, then we can help buddhababy002 with the problem if buddhababy002 still needs a hand! Haha!

- anonymous

let me check my text book

- anonymous

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)

- theEric

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\[8!=8*7*6*5*4*3*2*1\]\[5!=5*4*3*2*1\]

- anonymous

how would i put that into a calculator?

- theEric

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?

- anonymous

okay.. let me check

- anonymous

yes i do have that key

- theEric

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

- anonymous

i get fact (14)
87178291200

- theEric

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?

- theEric

And do you want me to explain the formula, just for your own conceptual understanding?

- anonymous

let me try the problem

- theEric

Okay.

- anonymous

i got 120

- theEric

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

- theEric

Actually, calculator.

- theEric

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!.

- theEric

87178291200 / 120

- theEric

Hmm.. I see a 120! Haha!

- anonymous

i got it.. lols thanks its so complicating

- theEric

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?

- theEric

http://www.mathsisfun.com/combinatorics/combinations-permutations.html
seems to be helpful in understanding combinations and permutations.

- theEric

I can also attempt to explain, if you'd like!

- theEric

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!

- theEric

Maybe I could explain it well now!

- anonymous

i think i kinda of got it.. thanks

- anonymous

:D

- theEric

Alright! Good luck with everything! Thank you! :)

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