Question

An instructor gives an exam with fifteen questions. Students are allowed to choose any eleven to answer.

Required:
a. How many different choices of eleven questions are there?
b. Suppose seven questions require proof and nine do not. How many groups of eleven questions contain five that require proof and six that do not?

Answer 1

Answer:

a. There are 1365 choices of eleven questions.

b. 1764 groups of eleven questions contain five that require proof and six that do not.

Step-by-step explanation:

The order in which the questions are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

a. How many different choices of eleven questions are there?

Eleven questions from a set of 15. So

$C_{15,11} = \frac{15!}{11!4!} = 1365$

There are 1365 choices of eleven questions.

b. Suppose seven questions require proof and nine do not. How many groups of eleven questions contain five that require proof and six that do not?

5 from a set of 7 and 6 from a set of 9. So

$C_{7,5}C_{9,6} = \frac{7!}{5!2!} \times \frac{9!}{6!3!} = 1764$

1764 groups of eleven questions contain five that require proof and six that do not.