Posted by sam on Sunday, August 23, 2009 at 1:15pm.
Is this the right way to answer this question?
A physics exam consists of 9 multiplechoice questions and 6 openended problems in which all work must be shown. If an examinee must answer 7 of the multiplechoice questions and 3 of the openended problems, in how many ways can the questions and problems be chosen?
(9x8x7x6x5x4x3)x (6x5x4) = 21772800

math  Marth, Sunday, August 23, 2009 at 1:25pm
The number of possibilities is nCk, or n choose k.
nCk = n! / (k! (nk)!)
For example, the number of possibilities for the multiple choice questions is 9C7.
There are 2 types of questions  multiply the number of possibilities for each question. 
math  sam, Sunday, August 23, 2009 at 1:40pm
I tried that and it doesn't seem to add up.
n! = 9x8x7x6x5x4x3x2 / (k= 7x6x5x4x3x2(2) = 362880/5040(2)=362880/10080 = 36 there seems to be more possibilities than that 
math  Marth, Sunday, August 23, 2009 at 1:56pm
For the multiple choice questions, the examinee is choosing 7 out of 9 questions. There are not so many ways you can choose 7 of those 9  remember order does not matter.
Also, after you compute the 6C3 for the open ended, and multiply them, you end up with 720 possibilities total. 
math  sam, Sunday, August 23, 2009 at 2:11pm
Thank you thank you!! I understand now!