Question

A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the question?

Answer 1

`color(blue)(10)`

Explanation:

We can solve this using the number of combinations of `bbn` things taken `bbr` at a time.

When solving these types of problems, you need to determine whether you are looking for combinations or permutations.

A permutation is concerned with the elements in a set as well as the order in which they appear.

Example:

The number of permutations of `ABC` taken 3 at a time is 6.

These would be:

ABC

ACB

BAC

BCA

CAB

CBA

Notice that all contain the same elements, but they are arranged in a different order.

For the number of combinations of ABC taken 3 at a time, there is only 1. This is:

ABC

This is because combinations are only concerned with the elements themselves not in the order they are arranged

For our problem we need combinations, because we are not interested in the order the questions are in, only that they are different questions.

Combinations are represented by:

`color(white)(0)^nC_(r)`

Where `n` is the number of elements and `r` is how many at a time they are taken:

`color(white)(0)^nC_(r)=(n!)/(r!(n-r)!`

`n!` means `n(n-1)(n-2)...(n-n+2)(n-n+1)`

To make this clearer:

`4! =4xx3xx2xx1`

We have 5 things, so `n=5` and we are choosing 3 at a time, so `r=3`

Putting these in:

`(n!)/(r!(n-r)!`

`(5!)/(3!(5-3)!) =(5xx4xx3xx2xx1)/(3xx2xx1xx2xx1)`

`=(5xx4xxcancel(3)xxcancel(2)xxcancel(1))/(cancel(3)xxcancel(2)xx1xx2xxcancel(1))=(5xx4)/(1xx2)=20/2=10`

So there are `color(blue)(10)` ways of choosing the 3 questions.