Question

Please help! :(

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?

A. 726,485,760

B. 14

C. 2,002

D. 126

Answer 1

A

Answer 2

Thank you, that is what I thought it was as well.

Answer 3

Assuming the problem is asking how many different combinations of articles can the editor make, without caring about the ordering, the answer should be "14 choose 9" or \[\frac{14!}{9!(14-9)!} = \frac{14!}{9!*5!} =\]

Answer 4

Are you sure it's B?

Answer 5

Let's take a simple case: there are 3 articles: A, B, C and only space to publish 2. The possible combinations are:
AB AC BC
3 combinations. Let's test the formula:
\[\frac{3!}{2!*1!} = \frac{3*2*1}{2*1*1} = 3\]Checks out.

Did I say B? I said you have to evaluate "14 choose 9" and gave the formula...

Answer 6

\[14! = 14*13*12*11*10*9*8*7*6*5*4*3*2*1\]\[9! = 9*8*7*6*5*4*3*2*1\]\[5!=5*4*3*2*1\]Hopefully it is apparent that the 9! in the denominator cancels out much of 14! leaving just \[\frac{14*13*12*11*10}{5*4*3*2*1}=\]

Answer 7

A

Answer 8

A

Answer 9

The editor isn't interested in permutations, only combinations.