Chapter 6 Counting Methods and the Pigeonhole Principle Section one Basic Principles Multiplication Principle Fact: The mn (multiplication) rule
¾ m elements from group A
¾ n elements from group B
¾ Select one from each group. There are m × n pairs can be sleeted.
Examples:
1. How many ways can you choose a car from one of three styles and in one of four paint
colors?
Styles
Colors
1
2
1
3
4
1
2
2
3
4
1
2
3
3
4
2. How many simple events are in tossing 2 dice?
6 × 6 = 36
3. Two candies are selected one at a time from a jar that contains one yellow and two red
candies. How many simple events are there?
3× 2 = 6
4. [2 urns problem]
¾ Urn 1: 2 W, 1 B
¾ Urn 2: 1 W.
¾ Draw one ball from Urn 1 and put into Urn 2.
¾ Draw one ball from Urn 2.
3× 2 = 6
1
Fact: k groups extension
¾ n1 elements from group 1
¾ n 2 elements from group 2
¾ …
¾ n k elements from group k
Select one from each group. There are
n1 × n 2 × n3 × ⋅ ⋅ ⋅ × n k ways can be sleeted.
Example 1: [Toss 3 coins.] How many simple events? [8]
Example 2: Truck driver, from A -> B -> C -> D.
A -> B
3
B -> C
4
C -> D
3
3 × 4 × 3 = 36.
How many ways from A -> D?
Example 3: There are 4 letters, A, B, C, D.
¾ Take two and permute them in a row:
4 × 3 = 12 ways
¾ Take three and permute them in a row:
4 × 3 × 2 = 24 ways
Permutation: In general,
n distinct objects to be taken r and permute them in a row:
There are n × ( n − 1) × ( n − 2) × ⋅ ⋅ ⋅ × ( n − r + 1) ways to select them.
If r = n , there are
n × ( n − 1) × ( n − 2) × ⋅ ⋅ ⋅ × 1 = n! ways to select them.
Addition Principle Let , ,
be sets and | |
, 1
. Assume that if the number of possible elements that can be selected from or
then
or or
. Then, is Example on Page 271 6·5·4
2
120 5·4
40, that is, the additional principle applied: 5 · 4
5·4
5·4
5·4
3·2·4
24 5·4
60 or 3 · 5 · 4
60 2
40 Inclusion‐Exclusion Principle for Two Sets: |
Theorem: If and are finite sets, then |
| |
| |
|
|. Example: | | 3 · 5 · 4 60 | | 3 · 5 · 4 60 The number of selections in which both Alice and Dolph are officers is |
| 3 · 2 · 4 24 Example: 1. How many eight bit strings begin with 111? 2 · 2 · 2 · 2 · 2 32 2. How many eight bit strings end with 00? 2 · 2 · 2 · 2 · 2 · 2 64 3. How many eight bit strings either begin with 111 or end with 00? The number of eight bit strings with beginning 111 and ending 00 is 2 · 2 · 2 8 The answer for this problem is 32 64 8 88 3