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Unformatted text preview: 2 = 16,777,214.
Continuing in this fashion we find the answer is: 3,737,091,842 (3.7 billion IP addresses)
12 Subtraction Rule
If a task can be done in either n1 ways or n2 ways,
then the number of ways to do the task is n1+n2
minus the number of ways to do the task that is
common to the two different ways.
Principle of InclusionandExclusion:
Let S1 and S2 be sets. Then
S1 ∪ S2 = S1+S2  S1 ∩ S2
13 Subtraction Rule: Example 1
How many bit strings of length 8 either start with a 1 bit or end
with the last two bits equal to 00 ?
Let S1 be the set of bit strings of length 8 that start with 1.
Then S1 = 27=128.
Let S2 be the set of bit strings of length 8 that end with 00.
Then S2=26=64.
Furthermore, the set of bit strings of length 8 that start with
bit 1 and end with bits 00 has cardinality S1 ∩ S2 = 25=32.
Thus, the answer is S1 ∪ S2=S1+S2S1 ∩ S2=128+6432=160.
14 Subtraction Rule: Example 2
Let us consider some (slightly simplified) rules for passwords:
Passwords must be 2 characters long.
Each character must be
a) a letter [az],
b) a digit [09], or
c) one of the 10 special characters [!@#$%^&*()].
Each password must contain at least 1 digit or special
character. 15 Subtraction Rule: Example 2 (Cont.)
A legal password has a digit or a special character in position 1
or position 2.
These cases overlap, so the subtraction rule applies.
(# of passwords with valid symbol in position #1)
= (10+10)·(10+10+26)=20·46
(# of passwords with valid symbol in position #2) = 20·46
(# of passwords with valid symbols in both places): 20·20
Answer: 920+920−400 = 1,440
16 Pigeonhole Principle If k+1 objects are assigned to k places, then at
least one place must be assigned at least two
objects. 17 Generalized Pigeonhole Principle
Theorem: If N > k objects are assigned to k places, then at
least one place must be assigned at least ȺN/kȺ objects.
Proof: Seeking a contradiction, suppose every place has less
than ȺN/kȺ objects; so at most ≤ ȺN/kȺ−1 objects per place.
Then the total number of objects is at most k N
k
N
−1 <k
+ 1 − 1 = k (N/k ) = N
k Thus, there...
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 Fall '11
 math

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