1. CSCI 115 Chapter 3 Counting

2. CSCI 115 §3.1 Permutations

3. §3.1 – Permutations Theorem 3.1.1 – Multiplication Principle of Counting –Suppose 2 tasks t 1 and t 2 are to be performed in sequence. If t 1 can be performed in n 1 ways, and t 2 can be performed in n 2 ways, then the sequence of tasks t 1 t 2 can be performed in n 1 n 2 ways.

4. §3.1 – Permutations Theorem 3.1.2 –Suppose tasks t 1, t 2, …, t k are to be performed in sequence. If t 1 can be performed in n 1 ways, t 2 can be performed in n 2 ways and so on, then the sequence of tasks t 1 t 2 …t k can be performed in n 1 n 2 … n k ways.

5. §3.1 – Permutations Consider the following problem: –How many different sequences, each of length r, can be formed from the elements of a set A if: Elements can be repeated Elements in the sequence must be distinct –Number of permutations of n objects taken r at a time

6. §3.1 – Permutations Theorem 3.1.3 –Let A be a set with n elements, r  Z +, with 1  r  n. The number of sequences of length r that can be formed from elements of A, allowing repetitions, is n r.

7. §3.1 – Permutations Theorem 3.1.4 –Let A be a set with n elements, r  Z +, with 1  r  n. The number of permutations of n objects taken r at a time is: n. (n – 1). (n – 2)... (n – (r – 1)) and is denoted n P r. –Equivalently: n!. (n – r)! Factorials

8. CSCI 115 §3.2 Combinations

9. §3.2 – Combinations Consider the following problem: –Let A be any set with n elements, 1  r  n, r  Z +. How many different subsets of r elements are there? Number of combinations of n objects taken r at a time

10. §3.2 – Combinations Theorem 3.2.1 –Let A be a set with |A| = n. Let r  Z + with 1  r  n. The number of combinations of n objects taken r at a time is n!. r!(n – r)! and is denoted n C r.

11. CSCI 115 §3.4 Elements of Probability

12. §3.4 – Elements of Probability Deterministic Experiment –Outcome should not change Finding acceleration due to gravity Probabilistic Experiment –Outcome can change Rolling a die and recording the outcome We will be discussing probabilistic experiments

13. §3.4 – Elements of Probability Sample space Event –Subset of sample space –Certain event –Impossible Event Mutually exclusive events

14. §3.4 – Elements of Probability Notation - Assigning probabilities to events –P(E) Frequency –If you have n experiments and E occurs n E times, then: f E = n E /n is the Frequency of occurrence of E in n trials f E  P(E) as n 

15. §3.4 – Elements of Probability Probability Spaces Axioms for a probability space A: –P1: 0  P(E)  1  E  A –P2: P(A) = 1 and P(  ) = 0 –P3: If E 1, E 2, …, E k are all mutually exclusive, then: P(E 1  E 2  …  E k ) = P(E 1 ) + P(E 2 ) + … + P(E k )

16. §3.4 – Elements of Probability Probability Spaces –Elementary Events (x i ) –Elementary Probability (P i = P({x i })) All the x i are mutually exclusive, and we have: –EP1: 0  P i  1 –EP2: P 1 + P 2 + … + P n = 1

17. §3.4 – Elements of Probability Equally likely outcomes –|A| = n, and all elementary events are equally likely, then: If E = {x 1, x 2, …, x k } then P(E) = k/n or P(E) = |E|/|A|

18. §3.4 – Elements of Probability Expected value –Sum of the value of each outcome times its probability –A way to calculate the ‘average’ value

19. CSCI 115 §3.5 Recurrence Relations

20. §3.5 – Recurrence Relations Recall sequences –Recursive –Explicit Recurrence relation –When an equivalent explicit formula is needed, the recursive formula is called a recurrence relation

21. §3.5 – Recurrence Relations Backtracking –‘Track’ the equation ‘back’ to an explicit formula –Does not always work

22. §3.5 – Recurrence Relations Linear Homogeneity –A recurrence relation is linearly homogenous of degree k if: a n = r 1 a n-1 + r 2 a n-2 + … + r k a n-k r i is constant  i –A recurrence relation that is linearly homogenous of degree k has the following characteristic equation: x k = r 1 x k-1 + r 2 x k-2 + … + r k The characteristic equation plays a role in determining the explicit formula

23. §3.5 – Recurrence Relations Theorem 3.5.1 –If the characteristic equation x 2 – r 1 x – r 2 = 0 of the recurrence relation a n = r 1 a n-1 + r 2 a n-2 has 2 distinct roots s 1 and s 2, then a n = us 1 n + vs 2 n (where u and v depend on initial conditions) is the explicit formula for the sequence. –If the characteristic equation x 2 – r 1 x – r 2 = 0 of the recurrence relation a n = r 1 a n-1 + r 2 a n-2 has a single root s, then a n = us n + vns n (where u and v depend on initial conditions) is the explicit formula for the sequence.