THE NATURE OF COUNTING

Copyright © Cengage Learning. All rights reserved.

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Copyright © Cengage Learning. All rights reserved.

12.3 Counting without Counting

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Counting without Counting

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Example 1 – Classifying permutations and combinations

Classify the following as permutations, combinations, or neither.

a. The number of three-letter “words” that can be formed using the letters {m, a, t, h}

b. The number of ways you can change a $1 bill with 5 nickels, 12 dimes, and 6 quarters

c. The number of ways a five-card hand can be drawn from a deck of cards

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Example 1 – Classifying permutations and combinations

d. The number of different five-numeral combinations on a combination lock

e. The number of license plates possible in Florida (routine plates have three numerals followed by three letters)

Solution:

a. Permutation, since mat is different from tam.

b. Combination, since “2 quarters and 5 dimes” is the same as “5 dimes and 2 quarters.”

cont’d

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Example 1 – Solution

c. Combination, since the order in which you receive the cards is unimportant.

d. Permutation, since “5 to the L, 6 to the R, 3 to the L, . . .” is different from “6 to the R, 5 to the L, 3 to the L, . . . .” We should not be misled by everyday usage of the word combination.

We made a strict distinction between combination and permutation—one that is not made in everyday terminology. (The correct terminology would require that we call these “permutation locks.”)

cont’d

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Example 1 – Solution

e. Neither; even though the order in which the elements are arranged is important, this does not actually fit the definition of a permutation because the objects are separated into two categories (pigeonholes).

The arrangement of letters is a permutation and the arrangement of numerals is a permutation, but to count the actual number of arrangements for this problem would require permutations and the fundamental counting principle.

cont’d

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License Plate Problem

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License Plate Problem

States issue vehicle license plates, and as the population increases, new plates are designed with more numerals and letters.

For example, The state of California has some plates consisting of three letters followed by three digits.

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License Plate Problem

When they ran out of these possibilities, they began making license plates with three digits followed by three letters.

Most recently they have issued plates with one digit followed by three letters in turn followed by three more digits.

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Example 2 – Find the number of possible license plates

How many possibilities are there for each of the following license plate schemes?

a. Three letters followed by three digits

b. Three digits followed by three letters

c. One digit followed by three letters followed by three digits

Solution:

a. 26 26 26 10 10 10 = 17,576,000Fundamental countingprinciple

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Example 2 – Solution

b. This is the same as part a. However, if both schemes are in operation at the same time, then the effective number of possible plates is

2 17,576,000 = 35,152,000

c. The addition of one digit yields ten times the number of possibilities in part a:

10 17,576,000 = 175,760,000

cont’d

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Example 2 – Solution

Note:

A state, such as California, that decides to leave the old plates in use as they move through parts a, b, and c could have the following number of license plates:

17,576,000 + 17,576,000 + 175,760,000 = 210,912,000

Notice that the fundamental counting principle is not used for all counting problems.

cont’d

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Which Method?

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Which Method?

We have now looked at several counting schemes: tree diagrams, pigeonholes, the fundamental counting principle, permutations, distinguishable permutations, and combinations.

In practice, you will generally not be told what type of counting problem you are dealing with—you will need to decide.

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Which Method?

Table 12.1 should help with that decision. Remember, tree diagrams and pigeonholes are applications of the fundamental counting principle, so they are not listed separately in the table.

Table 12.1Counting Methods.

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Example 7 – Find the number of ways

A club with 42 members wants to elect a president, a vice president, and a treasurer.

From the remaining members, an advisory committee of five people is to be selected.

In how many ways can this be done?

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Example 7 – Solution

This is both a permutation and a combination problem, with the final result calculated by using the fundamental counting principle.

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Example 9 – Find the number of ways of obtaining a diamond

Find the number of ways of obtaining at least one diamond when drawing five cards from an ordinary deck of cards.

Solution:

This is very difficult if we proceed directly, but we can compute the number of ways of not drawing a diamond:

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Example 9 – Solution

There is a total of 2,598,960 possibilities, so the number of ways of drawing at least one diamond is

2,598,960 – 575,757 = 2,023,203

cont’d