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544  8 / Advanced Counting Techniques


                                                because each of the r tokens may be a $1 token, a $2 token, or a $5 token. Because any number
                                                of tokens may be inserted, the number of ways to produce r dollars using $1, $2, or $5 tokens,
                                                                                                                 r
                                                when the order in which the tokens are inserted matters, is the coefficient of x in

                                                                                                  1
                                                             2    5         2   5 2
                                                    1 + (x + x + x ) + (x + x + x ) + ··· =
                                                                                                         5
                                                                                                    2
                                                                                           1 − (x + x + x )
                                                                                                 1
                                                                                        =                ,
                                                                                                   2
                                                                                           1 − x − x − x 5
                                                where we have added the number of ways to insert 0 tokens, 1 token, 2 tokens, 3 tokens, and
                                                                                                           2
                                                so on, and where we have used the identity 1/(1 − x) = 1 + x + x + ··· with x replaced
                                                         2
                                                              5
                                                with x + x + x . For example, the number of ways to pay for an item costing $7 using $1, $2,
                                                                                                                         7
                                                and $5 tokens, when the order in which the tokens are used matters, is the coefficient of x in this
                                                expansion, which equals 26. [Hint: To see that this coefficient equals 26 requires the addition
                                                                                              5 k
                                                                                          2
                                                                   7
                                                of the coefficients of x in the expansions (x + x + x ) for 2 ≤ k ≤ 7. This can be done by
                                                hand with considerable computation, or a computer algebra system can be used.]  ▲
                                                    Example 13 shows the versatility of generating functions when used to solve problems with
                                                differing assumptions.
                                EXAMPLE 13      Use generating functions to find the number of k-combinations of a set with n elements.Assume
                                                that the binomial theorem has already been established.

                                                Solution: Each of the n elements in the set contributes the term (1 + x) to the generating function
                                                         n      k
                                                             a
                                                f(x) =   k = 0 k x . Here f(x) is the generating function for {a k }, where a k represents the
                                                number of k-combinations of a set with n elements. Hence,
                                                                 n
                                                    f(x) = (1 + x) .

                                                But by the binomial theorem, we have


                                                            n
                                                                n   k
                                                    f(x) =         x ,
                                                                k
                                                           k = 0
                                                where


                                                     n        n!
                                                        =           .
                                                     k     k!(n − k)!

                                                Hence, C(n, k), the number of k-combinations of a set with n elements, is

                                                       n!
                                                             .
                                                    k!(n − k)!                                                                 ▲

                                                Remark: We proved the binomial theorem in Section 6.4 using the formula for the number of
                                                r-combinations of a set with n elements. This example shows that the binomial theorem, which
                                                can be proved by mathematical induction, can be used to derive the formula for the number of
                                                r-combinations of a set with n elements.
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