11.5. A summary of this principle is now stated. The complement of a set FIGURE 8


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1 11.5 The complement of a set FIGRE 8 Counting Problems Involving Not and Or When counting the number of ways that an event can occur (or that a task can be done), it is sometimes easier to take an indirect approach. In this section we will discuss two indirect techniques: first, the complements principle, where we start by counting the ways an event would not occur, and then the additive principle, where we break an event into simpler component events, and count the number of ways one component or another would occur. Suppose is the set of all possible results of some type and is the set of all those results that satisfy a given condition. For any set S, the cardinal number of S (number of elements in S) is written ns, and the complement of S is written S. Then Figure 8 suggests that n n n, or n n n. summary of this principle is now stated.
2 11.5 Counting Problems Involving Not and Or 665 Complements Principle of Counting If is any set within the universal set, then n() n() n(). By this principle, 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. (s we have seen in Chapters 2 and 3, the arithmetic operation of subtraction corresponds to the set operation of complementation and to the logical connective of negation, that is not. ) EXMPLE 1 For the set S a, b, c, d, e, f, find the number of proper subsets. Recall that a proper subset of S is any subset with fewer than all six elements. Several subsets of different sizes would satisfy this condition. However, it is easier to consider the one subset that is not proper, namely S itself. From set theory, we know that set S has a total of subsets. Thus, from the complements principle, the number of proper subsets is In words, the number of subsets that are proper is the total number of subsets minus the number of subsets that are not proper. Consider the tossing of three fair coins. Since each coin will land either heads (h) or tails (t), the possible results can be listed as follows. hhh, hht, hth, thh, htt, tht, tth, ttt (Even without the listing, we could have concluded that there would be eight possibilities. There are two possible outcomes for each coin, so the fundamental counting principle gives ) Now suppose we wanted the number of ways of obtaining at least one head. In this case, at least one means one or two or three. But rather than dealing with all three cases, we can note that at least one is the opposite (or complement) of fewer than one (which is zero). Since there is only one way to get zero heads (ttt), and there are a total of eight possibilities (as shown above), the complements principle gives the number of ways of getting at least one head: (The number of outcomes that include at least one head is the total number of outcomes minus the number of outcomes that do not include at least one head.) We find that indirect counting methods can often be applied to problems involving at least, or at most, or less than, or more than. EXMPLE 2 If four fair coins are tossed, in how many ways can at least one tail be obtained? By the fundamental counting principle, different results are possible. Exactly one of these fails to satisfy the condition of at least one tail (namely, no tails, or hhhh). So our answer (from the complements principle) is
3 666 CHPTER 11 Counting Methods EXMPLE 3 Carol Britz and three friends are boarding an airliner just before departure time. There are only ten seats left, three of which are aisle seats. How many ways can the four people arrange themselves in available seats so that at least one of them sits on the aisle? The word arrange implies that order is important, so we shall use permutations. t least one aisle seat is the opposite (complement) of no aisle seats. The total number of ways to arrange four people among ten seats is P10, The number of ways to arrange four people among seven (nonaisle) seats is P7, Therefore, by the complements principle, the number of arrangements with at least one aisle seat is The complements principle is one way of counting indirectly. nother technique is related to Figure 9. If and B are disjoint sets (have no elements in common), then writing the union of sets and B as B gives the principle stated here. B Disjoint sets FIGRE 9 Special dditive Counting Principle If and B are disjoint sets, then n( B) n() n(b). (This principle is special because and B are disjoint. It will be generalized later.) The principle states that if two conditions cannot both be satisfied together, then the number of ways that one or the other of them could be satisfied is found by adding the number of ways one of them could be satisfied to the number of ways the other could be satisfied. The idea is that we can sometimes analyze a set of possibilities by breaking it into a set of simpler component parts. (The arithmetic operation of addition corresponds to the set operation of union and to the logical connective of disjunction, that is or. ) Results in Examples 3 and 4 are supported in this screen. EXMPLE 4 How many fivecard hands consist of either all clubs or all red cards? No hand that satisfies one of these conditions could also satisfy the other, so the two sets of possibilities (for all clubs and all red cards) are disjoint. Therefore, the special additive principle applies. Since there are 13 clubs and 26 red cards, we obtain C13, 5 C26, ,780 67,067. The special additive principle also extends to the union of three or more disjoint sets, as illustrated by the next example. EXMPLE 5 Peggy Jenders needs to take twelve more specific courses for a bachelors degree, including four in math, three in physics, three in computer science, and two in business. If five courses are randomly chosen from these twelve for next semester s program, how many of the possible selections would include at least two math courses? Of all the information given here, what is important is that there are four math courses and eight other courses to choose from, and that five of them are being
4 11.5 Counting Problems Involving Not and Or 667 B C T = B C FIGRE 10 B selected for next semester. If T denotes the set of selections that include at least two math courses, then we can write T B C where the set of selections with exactly two math courses, B the set of selections with exactly three math courses, and C the set of selections with exactly four math courses. (In this case, at least two means exactly two or exactly three or exactly four.) The situation is illustrated in Figure 10. By previous methods, we know that n C4, 2 C8, , nb C4, 3 C8, , and nc C4, 4 C8, , so that, by the additive principle, nt Figure 11 illustrates the case in which two sets and B are not disjoint. In this case, when the number of items in set is added to the number in set B, any items in the intersection, B, get counted twice. To correct for this double inclusion, that number of items must be subtracted from the sum. The result is the following general principle. (This formula was first seen in Chapter 2, where it was called the cardinal number formula.) Nondisjoint sets FIGRE 11 General dditive Counting Principle If and B are any two sets, disjoint or not, then n( B) n() n(b) n( B). EXMPLE 6 Table 8 categorizes a diplomatic delegation of 18 congressional members 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 Democrat or a woman? TBLE 8 Men (M) Women (W) Totals Republican (R) Democrat (D) Totals sing the general additive counting principle, we obtain nd W nd nw nd W
5 668 CHPTER 11 Counting Methods EXMPLE 7 How many threedigit counting numbers are multiples of 2 or multiples of 5? multiple of 2 must end in an even digit (0, 2, 4, 6, or 8), so there are threedigit multiples of 2. multiple of 5 must end in either 0 or 5, so there are of those. multiple of both 2 and 5 is a multiple of 10 and must end in 0. There are of those, giving possible threedigit numbers that are multiples of 2 or multiples of 5. EXMPLE 8 single card is drawn from a standard 52card deck. (a) In how many ways could it be a heart or a king? single card can be both a heart and a king (the king of hearts), so use the general additive formula. There are thirteen hearts, four kings, and one card that is both a heart and a king: (b) In how many ways could the card be a club or a face card? There are 13 clubs, 12 face cards, and 3 cards that are both clubs and face cards, giving EXMPLE 9 How many subsets of a 25element set have more than three elements? It would be a real job to count directly all subsets of size 4, 5, 6,..., 25. It is much easier to count those with three or fewer elements: There is C25, 0 1 size0 subset. There are C25, 1 25 size1 subsets. There are C25, size2 subsets. There are C25, size3 subsets. The total number of subsets (of all sizes, 0 through 25) is ,554,432 (use a calculator). So the number with more than three elements must be 33,554, ,554, In Example 9, we used both the special additive formula (to get the number of subsets with no more than three elements) and the complements principle. Problem Solving 33,551,806. s you work the exercises of this section, keep in mind that indirect methods may be best, and that you may also be able to use permutations, combinations, the fundamental counting principle, or listing procedures such as product tables or tree diagrams. lso, you may want to obtain combination values, when needed, from Pascal s triangle, or find combination and permutation values on a calculator.
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