Part 2' FURTHER TOPICS IN ENUMERATION

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Part 2. FURTHER TOPICS INENUMERATION

• Ch9. Generating Functions

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9.1 Introductory Examples

• c1 c2 c3 c4 25 with 0 ci lt 10 for all 1
  i 4.
• If, in addition, we wanted c2 to be even and c3
  to be a multiple of 3, we could apply the results
  of Chapters 1 and 8 to several subcases.
• The power of the generating function rests upon
  its ability not only to solve the kinds of
  problems we have considered so far but also to
  aid us in new situations where additional
  restrictions may be involved.

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EXAMPLE 9.1

• While shopping one Saturday, Mildred buys 12
  oranges for her children, Grace, Mary, and Frank.
  
• In how many ways can she distribute the oranges
  so that Grace gets at least four, and Mary and
  Frank get at least two, but Frank gets no more
  than five?
• We see that we have all the integer solutions to
  the equation c1 c2 c3 12 where 4 c1, 2
  c2, and 2 c3 5.

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• When multiplying polynomials we add the powers of
  the variable, and here, when we multiply the
  three polynomials,
• (x4 x5 x6 x7 x8)(x2 x3 x4 x5
  x6)(x2 x3 x4 x5), two of the ways to obtain
  x12 are as follows
• 1) From the product x4x3x5, where x4 is taken
  from (x4 x5 x6 x7 x8), x3 from (x2 x3
  x4 x5 x6), and x5 from (x2 x3 x4 x5).
• 2) From the product x4x4x4, where the first x4 is
  found in the first polynomial, the second x4 in
  the second polynomial, and the third x4 in the
  third polynomial.
• Examining the product (x4 x5 x6 x7 x8)(x2
  x3 x4 x5 x6)(x2 x3 x4 x5) more
  closely, we realize that we obtain the product
  xixj xk for every triple (i, j, k) that appears
  in Table 9.1.
• Consequently, the coefficient of x12 in
• f (x) (x4 x5 x6 x7 x8)(x2 x3 x4
  x5 x6)(x2 x3 x4 x5) counts the number of
  distributionsnamely, 14that we seek.
• The function f (x) is called a generating
  function for the distributions.

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• But where did the factors in this product come
  from?
• The factor x4 x5 x6 x7 x8, for example,
  indicates that we can give Grace 4 or 5 or 6 or 7
  or 8 of the oranges.
• Once again we make use of the interplay between
  the exclusive or and ordinary addition.
• The coefficient of each power of x is 1 because,
  considering the oranges as identical objects,
  there is only one way to give Grace four oranges,
  one way give her five oranges, and so on.
• Since Mary and Frank must each receive at least
  two oranges, the other terms (x2 x3 x4 x5
  x6) and (x2 x3 x4 x5) start with x2, and
  for Frank we stop at x5 so that he doesnt
  receive more than five oranges.
• Why does the term for Mary stop at x6?

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EXAMPLE 9.2

• If there is an unlimited number (or at least 24
  of each color) of red, green, white, and black
  jelly beans,
• in how many ways can Douglas select 24 of these
  candies so that he has an even number of white
  beans and at least six black ones?

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• red (green) 1 x x2 x24, where the
  leading 1 is for 1x0, because one possibility for
  the red (and green) jelly beans is that none of
  that color is selected
• white (1 x2 x4 x6 x24)
• black (x6 x7 x8 x24)
• So the answer to the problem is the coefficient
  of x24 in the generating function
• f (x) (1 x x2 x24)2(1 x2 x4
  x6 x24) (x6 x7 x8 x24)

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EXAMPLE 9.3

• How many integer solutions are there for the
  equation c1 c2 c3 c4 25 if 0 ci for all
  1 i 4?
• We can alternatively ask in how many ways 25
  (identical) pennies can be distributed among four
  children.
• For each child the possibilities can be described
  by the polynomial 1 x x2 x3 x25.
  Then the answer to this problem is the
  coefficient of x25 in the generating function f
  (x) (1 x x2 x25)4.
• The answer can also be obtained as the
  coefficient of x25 in the generating function
  g(x) (1 x x2 x3 x25 x26
  )4,
• Note that the terms xk, for all k 26, are never
  used. So why bother with them?
• Because there will be times when it is easier to
  compute with a power series than with a
  polynomial.

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Exercise 9.1

• 2
• 4

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9.2 Definition and ExamplesCalculational
Techniques

• In this section we shall examine a number of
  formulas and examples dealing with power series.
• These will be used to obtain the coefficients of
  particular terms in a generating function.

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Definition 9.1

• Let a0, a1, a2, . . . be a sequence of real
  numbers. The function
• is called the generating function for the given
  sequence.

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EXAMPLE 9.4
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EXAMPLE 9.5
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• b) Extending the idea in part (a), we find that
  1 (1 - x)(1 x x2 x3 x4 ),
• So 1/1 - x is the generating function for the
  sequence 1, 1, 1, 1, . . . .
• Note that 1/(1 - x) 1 x x2 x3 is
  valid for all real x where x lt 1
• it is for this set of values that the geometric
  series 1 x x2 x3 converges.
• In our work with generating functions we shall be
  primarily concerned with the coefficients of the
  powers of x.

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EXAMPLE 9.6

• a) Rewriting the result in part (b) of Example
  9.5, we have 1/(1 - y) 1 y y2 y3
  .
• Upon substituting 2x for y, we then learn that
  1/(1 - 2x) 1 (2x) (2x)2 (2x)3
  1 2x 22x2 23x3 ,
• so 1/(1 - 2x) is the generating function for the
  sequence 1 ( 20), 2 ( 21), 22, 23, . . . .
• In fact, for each a ? R, it follows that 1/(1 -
  ax) 1 (ax) (ax)2 (ax)3 1 ax
  a2x2 a3x3 ,
• so 1/(1 - ax) is the generating function for the
  sequence 1 ( a0), a ( a1), a2, a3, . . . .
• Here we want 00 1 for the case where a 0.

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EXAMPLE 9.7

• For n ? Z, the Maclaurin series expansion for (1
  x)-n is given by

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EXAMPLE 9.8

• Find the coefficient of x5 in (1 - 2x)-7.

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EXAMPLE 9.9
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EXAMPLE 9.11

• In how many ways can we select, with repetitions
  allowed, r objects from n distinct objects?
• For each of the n distinct objects, the geometric
  series 1 x x2 x3 represents the
  possible choices for that object (namely none,
  one, two, . . .) .
• Considering all of the n distinct objects, the
  generating function is f (x) (1 x x2 x3
  )n, and the required answer is the
  coefficient of xr in f (x). Now

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EXAMPLE 9.14

• In how many ways can a police captain distribute
  24 rifle shells to four police officers so that
  each officer gets at least three shells, but not
  more than eight?
• f (x) (x3 x4 x8)4.
• We seek the coefficient of x24 in f (x). With (x3
  x4 x8)4
• x12(1 x x2 x5)4
• x12((1 - x6)/(1 - x))4, the answer is the
  coefficient of x12 in (1 - x6)4 (1 - x)-4

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EXAMPLE 9.15

• Since (1 x)2n (1 x)n2, by comparison of
  coefficients (of like powers of x), the
  coefficient of xn in (1 x)2n, which is C(2n, n)
  , must equal the coefficient of xn in

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EXAMPLE 9.15 (Skip)

• Since 1/(x - a) (-1/a)(1/(1 - (x/a)))
  (-1/a)1 (x/a) (x/a)2 for any a ? 0,
  
• we could solve this problem by finding the
  coefficient of x8 in

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EXAMPLE 9.17

• Use generating functions to determine how many
  four-element subsets of S 1, 2, 3, . . . , 15
  contain no consecutive integers.

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Sol (a) of EXAMPLE 9.17

• Consider one such subset (say 1, 3, 7, 10), and
  write 1 1 lt 3 lt 7 lt 10 15.
• We see that this set of inequalities determines
  the differences 1 - 1 0, 3 - 1 2, 7 - 3 4,
  10 - 7 3, and 15 - 10 5, and these
  differences sum to 14.
• Considering another such subsetsay 2, 5, 11,
  15, we write 1 2 lt 5 lt 11 lt 15 15 these
• inequalities yield the differences 1, 3, 6, 4,
  and 0, which also sum to 14.
• These examples suggest a one-to-one
  correspondence between the four-element subsets
  to be counted and the integer solutions to c1
  c2 c3 c4 c5 14 where 0 c1, c5, and 2
  c2, c3, c4. (Note Here c2, c3, c4 2 guarantee
  that there are no consecutive integers in the
  subset.)
• The answer is the coefficient of x14 in f (x)
  (1 x x2 x3 )(x2 x3 x4
  )3(1 x x2 x3 ) x6(1 - x)-5.

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EXAMPLE 9.19

• Let f (x) x/(1 - x)2. This is the generating
  function for the sequence a0, a1, a2, . . . ,
  where ak k for all k ? N.
• The function g(x) x(x 1)/(1 - x)3 generates
  the sequence b0, b1, b2, . . . , for bk k2, k
  ? N.
• The function h(x) f (x)g(x) consequently gives
  us a0b0 (a0b1 a1b0)x (a0b2 a1b1 a2b0)x2
  , so h(x) is the generating function for
  the sequence c0, c1, c2, . . . , where for each k
  ? N,
• ck a0bk a1bk-1 a2bk-2 ak-2b2
  ak-1b1 akb0.

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• Here, for example, we find that
• c0 0 02 0
• c1 0 12 1 02 0
• c2 0 22 1 12 2 02 1
• c3 0 32 1 22 2 12 3 02 6

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• Whenever a sequence c0, c1, c2, . . . arises from
  two generating functions f (x) for a0, a1, a2, .
  . . and g(x) for b0, b1, b2, . . ., as in this
  example, the sequence c0, c1, c2, . . . is called
  the convolution of the sequences a0, a1, a2, . .
  . and b0, b1, b2, . . . .

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EXAMPLE 9.20

• For f (x) 1/(1 - x) 1 x x2 x3
  and g(x) 1/(1 x) 1 - x x2 - x3 ,
• we find that f (x)g(x) 1/(1 - x)(1 x)
  1/(1 - x2) 1 x2 x4 x6 .
• Consequently, the sequence 1, 0, 1, 0, 1, 0, . .
  . is the convolution of the sequences 1, 1, 1,1,
  1, 1, . . . and 1, -1, 1, -1, 1, -1, . . . .

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EXAMPLE 9.2 (modified)

• If there is an unlimited number (or at least 24
  of each color) of red, green, white, and black
  jelly beans,
• in how many ways can Douglas select 24 of these
  candies so that he has an even number of white
  and green beans?

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• red 1 x x2 , where the leading 1 is
  for 1x0, because one possibility for the red (and
  green) jelly beans is that none of that color is
  selected
• White (green) (1 x2 x4 x6 )
• So the answer to the problem is the coefficient
  of x24 in the generating function
• f (x) (1 x x2 )2 (1 x2 x4 x6
  )2

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Exercise 9.2

• 2 c), e)
• 6
• 10
• 14 (hint ???????sample space)
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