SOLUTION: Club members are going to elect their officers. If there are four candidates for president, 3 for vice president, and 2 for secretary, then how many ways can the officers be el

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Question 1193984: Club members are going to elect their officers. If there are four candidates for
president, 3 for vice president, and 2 for secretary, then how many ways can the
officers be elected?

Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 24

Reason:
Multiply the values given
4*3*2 = 24

Why does this work?
Well let's consider just the president and VP for now.
We have 4 people for president and 3 for VP.
Make a table with 4 rows and 3 columns.
The rows correspond to the candidates for president, while the columns are for VP.
This table has 4*3 = 12 inner cells to show the twelve different president/VP combos.

To further extend this thought exercise, now consider a table with 12 rows and 2 columns.
The rows represent all the possible president/VP combos covered in the previous paragraph.
The columns represent the different possibilities for secretary.
This second table has 12*2 = 24 different cells. Each of the 24 cells shows a different combo of president/VP/secretary.

We can rewrite 12*2 = 24 into 4*3*2 = 24 to see why we multiply the values.

Side note: Optionally you could create a tree diagram to help see all the possible outcomes.

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.

In   4*3*2 = 24   different ways.

Refer to  Fundamental  Counting principle.

To see many other similar solved problems of this kind,  look into the lesson
    - Fundamental counting principle problems
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.