12.6. Counting with Permutations and Combinations http://www.ck12.org
are four options for the first passenger seat. Once that person is seated there are three options for the next passenger
seat. This goes on until there is one person left with one seat.
1 · 4 · 3 · 2 · 1 = 24
Example B
How many different ways can the gold, silver and bronze medals be awarded in an Olympic event with 12 athletes
competing?
Solution:Since the order does matter with the three medals, this is a permutation problem. You will start with 12
athletes and then choose and arrange 3 different winners.
12 P 3 =( 12 12!− 3 )!=12!9!=^12 ·^119 ··^10 ...·^9 ·...=^12 ·^11 ·^10 =^1320
Note that you could also use a decision chart to decide how many possibilities are there for gold (12) how many
possibilities are there for silver (11 since one already has gold) and how many possibilities are there for bronze (10).
You can use a decision chart for any permutation problem.
12 · 11 · 10 = 1320
Example C
You are deciding which awards you are going to display in your room. You have 8 awards, but you only have room
to display 4 awards. Right now you are not worrying about how to arrange the awards so the order does not matter.
In how many ways could you choose the 4 awards to display?
Solution:Since order does not matter, this is a combination problem. You start with 8 awards and then choose 4.
8 C 4 =
( 8
4
)
=4!( 8 8!− 4 )!= 48 ·· 37 ··^62 ··^51 = 7 · 2 · 5 = 70
Note that if you try to use a decision chart with this question, you will need to do an extra step of reasoning. There
are 8 options I could choose first, then 7 left, then 6 and lastly 5.
8 · 7 · 6 · 5 = 1680
This number is so big because it takes into account order, which you don’t care about. It is the same result you
would get if you used the permutation formula instead of the combination formula. To get the right answer, you
need to divide this number by the number of ways 4 objects can be arranged, which is 4!=24. This has to do with
the connection between the combination formula and the permutation formula.
Concept Problem Revisited
A license plate that has 3 letters and 4 numbers can be represented by a decision chart with seven spaces. You can
use a decision chart because order definitely does matter with license plates. The first spot is a number, the next three
spots are letters and the last three spots are numbers. Note that when choosing a license plate, repetition is allowed.
10 · 26 · 26 · 26 · 10 · 10 · 10 = 263 · 104 = 175 , 760 , 000
This number is only approximate because in reality there are certain letter and number combinations that are not
allowed, some license plates have extra symbols, and some commercial and government license plates have more
numbers, fewer letters or blank spaces.
Vocabulary
Acombinationis the number of ways of choosingkobjects from a total ofnobjects (order does not matter).
Apermutationis the number of ways of choosing and arrangingkobjects from a total ofnobjects (order does
matter)