uni/year2/semester1/logseq-stuff/pages/Advanced PIE, Derangements, & Counting Functions.md

• #MA284 - Discrete Mathematics
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• Advanced Counting Using PIE

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  • The PIE works for larger number of sets than 2 and 3, although it gets a little messy to write down.
    • For 4 sets, we can think of it as:
      • |A \cup B \cup C \cup D| = (the sum of the sizes of each single set) - (the sum of the sizes of each intersection of 2 sets) + (the sum of the sizes of each intersection of 3 sets) - (the sum of the sizes of the intersection of all 4 sets).

    • Example

      • How many ways can we distribute 10 slices of pie to 4 kids that such that no kid gets more than 2 slices (and each slice is distributed)? [See the textbook for a more detailed solution.]
        • The answer is obviously 0 - there will be 2 slices leftover after each kid gets the maximum of 2 slices.
        • Without the restriction that nobody gets more than 2 slices, there would be \binom{13}{3} = 286 ways of sharing distributing the slices (10+4-1 stars & 4-1 bars).
        • Now, count the number of ways where a child gets more than 2 slices, i.e. some child gets \geq 3 slices.
          • \binom{4}{1} \binom{7+3}{3} = 4(120) = 480
          • (choose one of 4 kids)(number of ways of distributing).
        • Add back in the doubly counted ones, subtracted the triply counted,
        • • \binom{13}{3} - \binom{4}{1} \binom{10}{3} + \binom{4}{2} \binom{7}{3} - \binom{4}{3} \binom{4}{3} + \binom{4}{4} \binom{1}{3} \\ = 286 - 480 +210 -16 = 0

      • Example

        • Not all problems have such easy solutions.
        • How many non-negative integer solutions are there to x_1 + x_2 + x_3 +x_4 +x_5 = 13 if:
          • 1. There are no restrictions (other than x_i being an nni).
            2. 0\leq x_i \leq 3 for each i.
          • 1. \displaystyle \binom{13+4}{4} = \binom{17}{4}
          • 2. Idea: All possibilities - "the wrong ones", i.e., count the possibilities where at least one of the x_i \geq 4.
            • \binom{5}{1} ways pf choosing the x_i and then number of solutions to x_1+x_2+x_3+x_4+x_5 = 9 is \binom{9+4}{4} = \binom{13}{4}, i.e. \binom{5}{1} \binom{13}{4}.
            • But we have double counted, so number of solutions with two x_i \ geq 4 is \binom{5}{2} choices and x_1+x_2+x_3+x_4+x_5 = 5 has \binom{5+4}{4} = \binom{9}{4} solutions.
            • Answer: \displaystyle \binom{17}{4} - \binom{5}{1} \binom{13}{4} + \binom{5}{2}\binom{9}{4} - \binom{5}{3}\binom{5}{4} +0.

• Derangements

  • What is a derangement? #card card-last-interval:: 3.45 card-repeats:: 2 card-ease-factor:: 2.46 card-next-schedule:: 2022-11-20T19:47:44.301Z card-last-reviewed:: 2022-11-17T09:47:44.301Z card-last-score:: 5
    • A derangement is a permutation where no element is left in its original place, everything is moved.

  • Example - Derangements of 4 Letters \text{STARS}.

    • Let D_n be the number of derangements of n objects.
    • First, we will work out the formulae for D_1, D_2, D_3, & D_4.
      • D_1 = 0,\ D_2 = 1,\ D_3 = 2,\ D_4 = 9
    • We derive a formula using PIE.
    • We know that there are 4! permutations. Which ones are not derangements?
      • Suppose that one item (at least) is left in place.
        • There are \displaystyle \binom{4}{1} \cdot 3! such permutations.
          • (choose one item to not change from four)(number of ways of permutating the other items).
          • However, some of these will be counted twice.
            • So, by PIE, the answer is
              • D_4 = 4! - \binom{4}{1}3! + \binom{4}{2}2!-\binom{4}{3}1!+\binom{4}{4}0!
              • D_4 = 4! - \frac{4!3!}{1!3!} +\frac{4!2!}{2!2!}-\frac{4!1!}{3!1!} + \frac{4!0!}{4!0!}
              • D_4 = 4![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}] = 9
    • In general, the formula for D_n, the number of derangements of n objects is
      • D_n = n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+ \dots + (-1)^n \frac{1}{n!})
    • Note that the series expansion for e^x is
      • e^x = 1 + \frac{x}{1!} +\frac{x^2}{2!}+\frac{x^3}{3!} + \dots
      • So \displaystyle e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!}-\frac{1}{3!}+ \dots
        • So \displaystyle \lim_{n \to \infty} \frac{D_n}{n!} = e^{-1} \approx 0.36787

• Counting with Repetitions

  • What is a Multinomial Coefficient? #card card-last-interval:: -1 card-repeats:: 1 card-ease-factor:: 2.5 card-next-schedule:: 2022-11-15T00:00:00.000Z card-last-reviewed:: 2022-11-14T20:07:58.830Z card-last-score:: 1
    • The number of different permutations of n objects, where there are n_1 indistinguishable objects of Type 1, n_2 indistinguishable objects of Type 2, ..., and n_k indistinguishable objects of Type k, is
      • \frac{n!}{(n_1!)(n_2!) \dots (n_k!)}

  • Example

    • How many "words" can we make from the letters in the set \{R,O,S,C,O,M,M,O,N\}.
      • If somehow the three $O$s were all distinguishable, and the two $M$s were distinguishable, the answer would be 9!.
    • But, since we can't distinguish the identical letters,
      • Let's choose which of the 9 positions in which we place the three $O$s.
        • This can be done in \displaystyle \binom{9}{3} ways.
      • Now, let's choose which of the remaining 6 positions in which we place the two $M$s.
        • This can be done in \displaystyle \binom{6}{2} ways.
      • Finally, let's choose where to replace the remaining 4 letters.
        • This can be done in 4! ways.
      • By the Multiplicative Principle, the answer is
        • \binom{9}{3}\binom{6}{2}4! = \frac{9!}{3!6!} \frac{6!}{2!}{4!} 4! = \frac{9!}{3!2!}

• Example (MA284 Semester 1 Exam, 2014/2015)

  • (i) Find the number of different arrangements of the letters in the place name WOLLONGONG.
    • OOOLLNNGGW
    • \frac{10!}{3!2!2!2!1!} = 75600
  • (ii) How many of these arrangements start with three Os?
    • OOO (one way) and 7 others.
    • \frac{7!}{2!2!2!} = 630
  • (ii) How many contain the two Gs consecutively?
    • Treat GG as a single letter and permute 9 letters.
    • \frac{9!}{3!2!2!1!} = 15120
  • (iv) How many do not contain the two Gs consecutively?
    • Use (i) - (iv).
    • 75600 - 15120 = 60480

• Counting Functions

  • Recall that f: A \rightarrow B is a function that maps every element of the set A onto some element of set B.
    • We call A the domain & B the codomain.
    • Each element of A gets mapped to exactly one element of B.
  • What does it mean if a is the image of b? #card card-last-interval:: -1 card-repeats:: 1 card-ease-factor:: 2.36 card-next-schedule:: 2022-11-23T00:00:00.000Z card-last-reviewed:: 2022-11-22T13:40:10.499Z card-last-score:: 1
    • If f(a) = b where a \in A and b \in B, we say that "the image of a is $b$", or, equivalently, "b is the image of $a$".
  • What is a surjective function (surjection)? #card card-last-interval:: -1 card-repeats:: 1 card-ease-factor:: 2.5 card-next-schedule:: 2022-11-15T00:00:00.000Z card-last-reviewed:: 2022-11-14T20:13:44.812Z card-last-score:: 1
    • For some function f: A \rightarrow B, if every element of B is the image of some element A, we say that the function is surjective (also called "onto").
  • What is an injective function (injection)? #card card-last-interval:: -1 card-repeats:: 1 card-ease-factor:: 2.5 card-next-schedule:: 2022-11-15T00:00:00.000Z card-last-reviewed:: 2022-11-14T20:16:06.397Z card-last-score:: 1
    • For some function f: A \rightarrow B, if no two elements of A have the same image in B, we say that the function is injective (also called "one-to-one").
  • What is a bijective function (bijection)? #card card-last-interval:: 4 card-repeats:: 2 card-ease-factor:: 2.7 card-next-schedule:: 2022-11-18T20:09:58.766Z card-last-reviewed:: 2022-11-14T20:09:58.766Z card-last-score:: 5
    • The function f: A \rightarrow B is a bijection if it is both surjective & injective.
      • Then f defines a one-to-one correspondence between A & B

  • Examples

    • Let A & B be finite sets. How many functions f: A \rightarrow B are there?
      • We can use ((6336be87-7dea-4ba3-b7d0-c77a73bae948)) to deduce that there are in total |B|^{|A|} functions from A to B.
    • How many functions f: A\{1,2,3,4,5,6,7,8\} \rightarrow \{1,2,3,4,5,6,7,8\} are bijective?
      • Remember what it means for a function to be bijective: ^^each element in the codomain must be the image of exactly one element of the domain.^^
      • What we are really doing is just rearranging the elements of the codomain, so we are defining a permutation of 8 elements.
        • Therefore, the answer to our question is 8!.
      • More generally, there are n! bijections of the set \{1,2,\cdots, n\} onto itself.
      • 2022年10月19日