Question

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?

Answer 1

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!

Answer 2

i think you would use combination

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

whats with the exclamation mark?

Answer 7

exclamation mark means factorial

Answer 8

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

Answer 9

because you would still pick out the same magazines ..

Answer 10

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

Answer 11

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!

Answer 12

14C9 @lgbasallote

Answer 13

@alexwee123 doesn't "assemble" mean arranging?

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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.

Answer 18

assemble really feels like an arrangement...

Answer 19

i hate these fancy words =_=

Answer 20

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

Answer 21

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...

Answer 22

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

Answer 23

let me check my text book

Answer 24

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)

Answer 25

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

Answer 26

how would i put that into a calculator?

Answer 27

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?

Answer 28

okay.. let me check

Answer 29

yes i do have that key

Answer 30

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

Answer 31

i get fact (14)
87178291200

Answer 32

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?

Answer 33

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

Answer 34

let me try the problem

Answer 35

Okay.

Answer 36

i got 120

Answer 37

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

Answer 38

Actually, calculator.

Answer 39

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!.

Answer 40

87178291200 / 120

Answer 41

Hmm.. I see a 120! Haha!

Answer 42

i got it.. lols thanks its so complicating

Answer 43

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?

Answer 44

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

Answer 45

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

Answer 46

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!

Answer 47

Maybe I could explain it well now!

Answer 48

i think i kinda of got it.. thanks

Answer 49

:D

Answer 50

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