posted by sam on .
Is this the right way to answer this question?
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen?
(9x8x7x6x5x4x3)x (6x5x4) = 21772800
The number of possibilities is nCk, or n choose k.
nCk = n! / (k! (n-k)!)
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.
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
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.
Thank you thank you!! I understand now!