# A professor gives his students 7 essay questions to prepare for an exam. Only 5 of the questions will actually appear on the exam. How many different exams are possible?

Feb 24, 2016

P(7, 5) = \frac{7!}{(7-5)!}=2520 ways.

#### Explanation:

Of the 5 questions, the first question can be chosen in 7 ways. Having chosen one question out of 7, the second question can be chosen in 6 ways, third question in 5 ways, the fourth question in 4 ways and the fifth question in 3 ways.
So the total number of ways of choosing 5 questions from 7 is 7\times6\times5\times4\times3=(7!)/((7-5)!).

This can be generalized as follows:

Permutations: The number of ways of choosing $r$ objects from a collection of $N$ objects is P(N, r)\equiv \frac{N!}{(N-r)!}