Accounting Final

Topics: Combination, Permutation, Binomial coefficient Pages: 15 (3525 words) Published: September 26, 2010
Section 5
Permutations and Combinations

In preceding sections we have solved a variety of counting problems using Venn diagrams and the generalized multiplication principle. Let us now turn our attention to two types of counting problems that occur very frequently and that can be solved using formulas derived from the generalized multiplication principle. These problems involve what are called permutations and combinations, which are particular types of arrangements of elements of a set. The sorts of arrangements we have in mind are illustrated in two problems:

Problem A How many words (by which we mean strings of letters) of two distinct letters can be formed from the letters {a, b, c}?

Problem B A construction crew has three members. A team of two must be chosen for a particular job. In how many ways can the team be chosen?

Each of the two problems can be solved by enumerating all possibilities.

Solution of Problem A There are six possible words, namely

abacbabccacb.

Solution of Problem B Designate the three crew members by a, b, and c. Then there are three possible two- person teams, namely

abacbc.

(Note that ba, the team consisting of b and a, is the same as the team ab.)

We deliberately set up both problems using the same letters in order to facilitate comparison. Both problems are concerned with counting the numbers of arrangements of the elements of the set {a, b, c}, taken two at a time, without allowing repetition (for example, aa was not allowed). However, in Problem A the order of the arrangement mattered, whereas in Problem B it did not. Arrangements of the sort considered in Problem A are called permutations, whereas those in Problem B are called combinations.

More precisely, suppose that we are given a set of n objects1.

Then a permutation of n objects taken r at a time is an arrangement of r of the n objects in a specific order. So, for example, Problem A was concerned with permutations of the three objects, a, b, c (n = 3), taken two at a time (r = 2).

A combination of n objects taken r at a time is a selection of r objects from among the n, with order disregarded. Thus, for example, in Problem B we considered combinations of the three objects a, b, c (n = 3), taken two at a time (r = 2).

1All are assumed to be different.

It is convenient to introduce the following notation for counting permutations and combinations. Let

P(n, r) = the number of permutations of n objects taken r at a time

C(n, r) = the number of combinations of n objects taken r at a time.

Thus, for example, from our solutions to Problems A and B, we have

P(3, 2) = 6C(3, 2) = 3.

Very simple formulas for P(n, r) and C(n, r) allow us to calculate these quantities for any n and r. Let us begin by stating the formula for P( n, r). For r = 1, 2, 3 we have, respectively,

|[pic] | | |[pic] |(two factors) | |[pic] |(three factors) |

and, in general,

| | |[pic] |

This formula is verified at the end of this section.

EXAMPLE 1 Applying the permutation formula Compute the following numbers.

(a) P(100, 2)(b) P(6, 4)(c) P(5, 5)

Solution (a) Here n = 100, r = 2. So we take the product of two factors, beginning with 100:

P(100, 2) = [pic] = 9900.

(b) P(6, 4) = 6 5 4 3 = 360

(c) P(5, 5) = 5 4 3 2 1 = 120

Now Try Exercise 1

In order to state the formula for C(n, r), we must introduce some further notation. Suppose that r is any positive...
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