Counting
If you go to Baskin-Robbins (31 flavors) and
make a triple-scoop ice-cream cone,
How many different arrangements can you
create (if you allow repeats)?
31.31.31 = 313 = 29791
If you don’t allow repeats?
31.30.29 = 26970
Counting
You need to visit 5 cities. How many possible
different orderings of visiting the cities (no
repeats) are there?
5  4  3  2 1  120
This calculation is common enough that we have a notation
for it – the factorial (symbolized by !)
5! 5  4  3  2 1  120
7! 7  6  5  4  3  2 1  5040
Permutations and Combinations
These deal with drawing items without replacement
Are different orderings counted separately?
In other words, is ABC considered different than
BAC?
If YES, we’re talking about Permutations
(arrangements)
If NO, we’re talking about Combinations
Permutations
How many different 4-number PIN numbers are
possible if no two numbers are the same?
10  9  8  7  5040
Notice
10! 10  9  8  7  6  5  4  3  2 1

 10  9  8  7
6!
6  5  4  3  2 1
Where’d the 6 come from? From 10-4 = 6
Permutations
The number of permutations (arrangements or
sequences) of r items selected from n
available items (without replacement) is:
n!
n Pr 
(n  r )!
Permutations
Suppose you have to pick 5 photos out of 15 for a
magazine layout. Ordering matters (different
layouts should be considered as separate). How
many possible layouts are there?
15!
15!

 360360
15 P5 
(15  5)! 10!
Combinations
The number of combinations of r items
selected from n different items (without
replacement) is:
n!
n Cr 
(n  r )! r!
The extra term in the denominator makes it so different
arrangements aren’t counted separately (ABC is considered
equivalent to CBA)
Combinations
Suppose you have to pick 4 of your 12 friends
to take on a free vacation you won. How
many possibilities are there?
12!
12!

 495
12 C4 
(12  4)!4! 8!4!
For comparison’s sake, the number of permutations would be
11880
Combinations for Probability
Suppose you have to pick 4 of your 12 friends to
take on a free vacation you won. You decide to
choose by pulling names from a hat. What is the
probability that Ann, Betty, Carlos, and Dean are
chosen?
That’s one choice out of all possibilities, so
1
P
495
Homework
3.7: 1, 3, 5, 9, 21, 25