Question

A class of 30 students must elect a representative to go to a local

conference. They must also elect one alternate representative, in case the
first representative becomes ill or unable to go at the last minute.
How many different ways can the class elect a representative and
a
alternate?

Answer 1

870.

Explanation:

There are 30 choices of representatives. This leaves 29 choices for alternate representatives.

If one person gets elected for representative, there are 29 choices for an alternate representative.

This is the case if any of the 30 students get elected as representatives.

That means you can set it up as:

x*y where x is the choice of representatives and y is the choice of alternate representatives.

30 * 29, or 870.

Here is a visual example. Students a - dd are shown here.

a b c d e f g h i j k l m n o p q r s t u v w x y z aa bb cc dd

In this case, a has been chosen as class president. Any of the other 29 students could be a alternate, and then paired with a. This makes 29 total possible pairings. This is then repeated for the possibility of every student becoming a president, which creates 30*29.

Please tell me if this is wrong, I will try to make changes!