http://www.ck12.org Chapter 2. Conditional Probability
Starting with 6, multiply the first 3 numbers of the factorial:
6 P 3 =^6 ·^5 ·^4 =^120
This represents the number of ways to arrange 3 objects that are chosen from a set of 6 different objects.
The formula to solve permutations like these is:
nPr=n!
(n−r)!Look at Example C above. In this example, the total number of objects(n)is 6, while the number of objects
chosen(r)is 3. We can use these 2 numbers to calculate the number of possible permutations (or the number of
arrangements) of 6 objects chosen 3 at a time.
nPr=n!
(n−r)!6 P 3 =6!
( 6 − 3 )!
6 P 3 =
6!
3!
=
6 × 5 × (^4) (×( 3 ×( 2 (×(( 1
3 × 2 × 1
6 P 3 =
120
1
6 P 3 =^120
Guided Practice
a. What is the total number of possible 4-letter arrangements of the letters ’s’, ’n’, ’o’, and ’w’ if each letter is used
only once in each arrangement?
b. A committee is to be formed with a president, a vice president, and a treasurer. If there are 10 people to select
from, how many committees are possible?
Answer:
a. In this problem, there are 4 letters to choose from, son=4. We want 4-letter arrangements; therefore, we are
choosing 4 objects at a time. In this example,r=4.