Solutions to Counting: Combinations and Permutations (6.1 – In class exercises)
1. (example) There is one way to order one object {  }:
1. 
(example) There are two ways to order two distinct objects { 
}:
1. 
2. 
a. There are six ways to order three distinct objects { 
List all six ways.
1. 
2. 
3.

4.
5.
6.
}.



b. How many ways are there to order four distinct objects { 
possible ordering in the space below.
}? Show every
You should have listed 4  3  2 1  24 different orderings.
c. How many ways can you order five distinct objects { 
distinct objects?
 }?
What about six
Five objects: 5  4  3  2 1  120 different orderings.
Six objects: 6  5  4  3  2 1  720 different orderings.
d. How about n distinct objects? Memorize this result.
n objects: n  (n  1)  (n  2)   2  1 , which is called n! for short. For example, the long
expression 6  5  4  3  2 1 is usually just called 6! for short.
Solutions to Counting: Combinations and Permutations (6.1 – In class exercises)
2. a.
b. 10  9  90 ways.
c. 10  9  8  720 ways.
d. Let us pretend there are 14 students in a class. There are 14 13  182 different ways to
have one student picked first and another student picked second. Of these 182 ways,
only one has me picked first and my best friend picked second. So the probability is
1/182 .
e. Think about how many you could pick 3 objects out of 8. It would be 8  7  6 . This is
kind of like 8! except a bunch of the lower numbers are missing. Here is a trick to write
it in a short way:
8  7  6  5  4  3  2  1 8!
8!
876 
 
5  4  3  2 1
5! (8  3)!
The advantage of this writing is that is in terms of the original numbers 3 and 8. By the
same reasoning, the number of ways to pick r of the objects out of a set of n objects can
be expressed as
n!
.
(n  r )!
3.
b. If the order of the officers mattered, it would be 10  9  90 ways. Since the order doesn’t
matter, 10  9  90 overcounts by a factor of 2 (every AB selection we want would be
counted as AB and BA in the overcount). So there are 45 ways.
c. If the order of the officers mattered, it would be 10  9  8  720 ways. Since the order
doesn’t matter, 10  9  8  720 overcounts by a factor of 6 (every ABC selection we want
would be counted as ABC, ACB, BAC, BCA, CAB, CBA in the overcount). So there
are 720/6 = 120 ways.
d.
Pr is always bigger than n Cr . How much bigger depends on r. Generally, n Pr will be
r! times larger than n Cr , because there are r! arrangements when order matters that
count as a single arrangements when order does not matter. Therefore,
n
n
Pr  r !  nCr or equivalently
Pr
 n Cr
r!
n