By Henry Sharp
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Extra resources for Finite functions: an introduction to combinatorial mathematics
Sample text
An , and An+1 be mutually disjoint sets. Thus, ∪ni=1 Ai and An+1 are disjoint sets. 6, we have that n | ∪n+1 i=1 Ai | = | ∪i=1 Ai | + |An+1 |. 30 2 Basic Counting Applying the inductive hypothesis yields n | ∪n+1 i=1 Ai | = | ∪i=1 Ai | + |An+1 | n n+1 |Ai | + |An+1 | = = i=1 |Ai |. i=1 Alternatively, if x ∈ Ai , then x ∈ / Aj for j = i. Therefore, x is counted once by / Ai for all i, then x is counted | ∪ni=1 Ai | and once by ni=1 |Ai |. Similarly, if x ∈ zero times by both sides of the equation.
If no one is in both clubs, how many different combinations of officers can be at the retreat? Solution Note there are P (10, 3) = 720 ways for the Science club to select its officers. Similarly, there are P (20, 5) = 1860480 ways for the Spanish club to select their officers. Hence by the Multiplication Principle, the number of combinations of officers is given by: P (10, 3)P (20, 5) = 720(1860480) = 1339545600. 3 The Mathletes Club has 15 members and is selecting five officers (president, vice-president, secretary, treasurer, and sergeant-at-arms).
Alternatively, there are nine choices for the first digit, namely anything but zero. There are then 106 choices for the remaining digits. Hence, there are 9 ∗ 106 = 9000000 telephone numbers that do not start with zero. (iii) If a number contains 911, then there are 104 choices for the remaining digits. There are five ways to place the sequence 911 within the number, so 5 ∗ 104 telephone numbers contain 911. Thus by the Subtraction Principle there are 107 − 5 ∗ 104 = 9950000 acceptable numbers.
Finite functions: an introduction to combinatorial mathematics by Henry Sharp
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