Probability question

Question

An exam has 39 questions on it and students need to answer 26 of the questions.  In how many ways can this be done if the first 5 questions and exactly 6 of the last 8 questions must be answered and questions #13 and #20 must not be answered because of typos?

Answer 1

First 5 questions must be answered: there is C(5,5) = 1 way to do this.

6 of last 8 questions must be answered: there is C(8,6) = 8*7/2 = 28 ways to do this.

26-11 = 15 questions to answer from a list of (39 - 5 - 8 - 2 = 24) questions; there is C(24,15) ways to this.

 

So altogether there are C(5,5)*C(8,6)*C(24,15) ways to do this exam.

 

Note C(n,m) is the binomial coefficient: C(n,m) = n!/(n-m)!/m!