Chapter 10 Counting Methods

Chapter 10: Counting Methods 10.1 Counting by Systematic Listing 10.2 Using the Fundamental Counting Principle 10.3 Using Permutations and Combinations 10.4 Using Pascal’s Triangle 10.5 Counting Problems Involving “Not” and “Or”

Counting Problems Involving “Not” and “Or” Section 10-5 Counting Problems Involving “Not” and “Or”

Counting Problems Involving “Not” and “Or” Understand the complements principle of counting. Apply the complements principle to a counting problem by considering the problem in terms of outcomes that do not occur (a “not” statement). Understand the additive principle of counting. Apply the additive principle to a counting problem by considering the problem in terms of the union of two outcomes (an “or” statement).

Counting Problems Involving “Not” and “Or” The counting techniques in this section, which can be thought of as indirect techniques, are based on some correspondences between set theory, logic, and arithmetic as shown on the next slide.

Set Theory/Logic/Arithmetic Correspondences Operation or Connective (Symbol) Complement Not Subtraction Union Or Addition

Problems Involving “Not” Suppose U is the set of all possible results of some type. Let A be the set of all those results that satisfy a given condition. The figure below suggests that U A

Complements Principle of Counting The number of ways a certain condition can be satisfied is the total number of possible results minus the number of ways the condition would not be satisfied. Symbolically, if A is any set within the universal set U, then

Example: Counting the Proper Subsets of a Set For the set S = {c, a, l, u, t, o, r}, find the number of proper subsets. Solution A proper subset has less than seven elements. Subsets of many sizes would satisfy this condition. It is easier to consider the one subset that is not proper, namely S itself. S has a total of 27 = 128 subsets. From the complements principle, the number of proper subsets is 128 – 1 = 127.

Example: Counting Coin-Tossing Results If five fair coins are tossed, in how many ways can at least one tail be obtained? Solution By the fundamental counting principle, there are 25 = 32 different results possible. Exactly one of these fails to satisfy “at least one tail.” So from the complements principle, we have the answer: 32 – 1 = 31.

Problems Involving “Or” Another technique to count indirectly is to count the elements of a set by breaking that set into simpler component parts. If the cardinal number formula (Section 2.4) says to find the number of elements in S by adding the number in A to the number in B. We must then subtract the number in the intersection if A and B are not disjoint. If A and B are disjoint, the subtraction is not necessary.

Additive Principle of Counting The number of ways that one or the other of two conditions could be satisfied is the number of ways one of them could be satisfied plus the number of ways the other could be satisfied minus the number of ways they could both be satisfied together. If A and B are any two sets, then If A and B are disjoint, then

Example: Counting Card Hands How many five-card hands consist of all hearts or all black cards? Solution The sets all hearts and all black cards are disjoint. n(all hearts or all black cards) = n(all hearts) + n(all black cards) = 13C5 + 26C5 = 1,287 + 65,780 = 67,067

Example: Counting Selections From a Diplomatic Delegation A diplomatic delegation of 20 congressional members are categorized as to political party and gender. If one of the members is chosen randomly to be spokesperson for the group, in how many ways could that person be a Republican or a man? Men (M) Women (W) Totals Republican (R) 8 4 12 Democrat (D) 3 5 11 9 20

Example: Counting Selections From a Diplomatic Delegation Solution (M) (W) Totals (R) 8 4 12 (D) 3 5 11 9 20

Example: Counting Three-Digit Numbers with Conditions How many three-digit counting numbers are multiples of 2 or multiples of 5? Solution There are 9(10)(5) = 450 three-digit multiples of 2. A multiple of 5 ends in a 0 or 5, so there are 9(10)(2) = 180 of those. A multiple of 2 and 5 must end in a 0. There are 9(10)(1) = 90 of those. So we have 450 + 180 – 90 = 540 three-digit counting numbers that are multiples of 2 or 5.

Example: Counting Card-Drawing Results A single card is drawn from a standard 52-card deck. a) In how many ways could it be a club or a queen? b) In how many ways could it be a red card or a face card? Solution a) club + queen – queen of clubs = 13 + 4 – 1 = 16 b) red card + face card – red face cards = 26 + 12 – 6 = 32