COMP 2711H: Lecture 6

Date: 2024-09-11 04:36:04

Reviewed:

Topic / Chapter: Counting

summary

❓Questions

Notes

Counting

Counting
- how many ways is there to do something?
PA: principle of addition
- there are 2 points, $a, b$
  - there exists 2 roads, 2 ferry routes, and 1 flight route
  - how many ways? $2 + 2 + 1 = 5$
- i.e. dividing big set's size into sum of smaller sets' size
PM: principle of multiplication
- when noe event is independent of another
- same city $a, b$
  - but intermediate $c, d$ also exists
  - a-c-d-b in linear order
  - a-c: 2; c-d: 3; d-b: 2 ways
  - total ways of a-b, without going back:
    - $2 \cdot 3 \cdot 2 = 12$ ways
- as long as no. of later choices (not individual cases of later choices) are independent of previous choices, we can simply multiply
- another way to divide a large set
Counting permutations
- prof's notation: $[n] = 1.. = n$
- permutations of ${1, 2, 3}$ :
  - ${1, 2, 3}, {1, 3, 1}, {2, 1, 3}, {2, 3, 1}, {3, 2, 1}, {3, 1, 2}$
  - 6 ways!
- how many permutation for $n$ numbers?
  - e.g. how many different (linear) queues?
    - use notation $x_{n}$ to denote permutation ways of $[n]$
  - $x_{1} = 1$
  - $x_{n} = x_{n - 1} \cdot n$
    - i.e. inserting $n$ -th person into queue of $n - 1$ people
    - $n$ slots to insert!
  - $x_{n} = \prod_{i = 1}^{n} i =: n!$
  - definition: can be either multiplication / permutation of $[n]$
    - $0! = 1$ as there is 1 way of aligning 10 people
- or, different approach:
  - for $n$ objects
    - how many choices do I have for 1st place?: $n$
    - how many choices do I have for 2nd place?: $n - 1$
    - ...
    - how many choices do I have for last place?: $1$
    - total: $n!$
  - 👨‍🏫 note that no. of later choices are independent of previous choices
Permutations
- how many queue of length $r$ can we form from $n$ people?
  - $\prod_{i = 0}^{r - 1} (n - i) =: P_{r}^{n} = P (n, r)$
  - or: $n! / (n - r)!$
  - 👨‍🏫: double counting: we consider $(a, b)$ and $(b, a)$ as different queue
    - order matters here!
- or, think of other way
  - it's still $n$ -length queue, but the last $n - r$ elements do not matter
    - thus: $n! / (n - r)!$
Permutations of non-distinct items
- how many permutations does each word have
  - not necessarily meaningful
  - $HK U ST$ : all letters are distinct
    - $5!$ ways to order
  - simple: $AA BB$
    - there are $4!$ ways to simply display
      - if it was $A_{1} A_{2} B_{1} B_{2}$
    - but order of 2 As and 2 Bs doesn't matter
      - so $2! 2!$ ways of double count
    - $\frac{4 !}{2 ! 2 !} = 6$ distinct ways
  - $A BR A C A D A BR A$ : some letters appear more than once!
    - 5 A, 2 B, 2 R, 1 C, 1 D
    - all permutations: $\frac{11 !}{5 ! 2 ! 2 !}$
- general way
  - there are $t$ alphabets in a word
  - where each letter appears $k_{t}$ times
  - total no. of permutations: $\frac{( \sum _{i = 1}^{t} k _{i} )!}{\prod _{i_{1}}^{t} k _{i} !}$
- one of the ways we show that a fraction is an integer
  - claim that it is answer to some counting problem
    - all counting problem: has integer solution
- 👨‍🏫 I'll fail you if you say there are 9.5 ways of counting!!!
Combinations
- 👨‍🏫 what if we don't care about the order?
  - e.g. how many subsets of $[n]$ have size $r$
  - notation: $C_{r}^{n}, C (n, r), (r n)$
- equivalent: how many binary seq. of length $n$ have $r$ ones?
  - $= \frac{n !}{( n - r )! r !}$
- or, a different way
  - $P (n, r)$ : no. of different permutation of length $r$ from $[n]$
    - then divide it by $r!$ : no. of different ordering of length $r$
  - thus: $\frac{P ( n , r )}{r !} = \frac{n !}{( n - r )! r !}$

Practice Problems (Permutations)

Idea
- $[n]$ has $2^{n}$ subsets
- old method we used:
- $0$ : false; $1$ : true
- and we can assign ${0, 1}$ for each $n$ elements
  - to assign: whether it's included in subset / not
- or, using counting:
  - there are 2 choices for first element
  - and so on for all $n$ elements, independent of previous choices
  - thus $2^{n}$ ways total
Problem 1
- how many subsets are there of $[n]$ , including no. $1$ ?
  - $2^{n - 1}$
  - consider: first element must be 1
    - only single way for first element
    - for the rest $n - 1$ elements: $2^{n - 1}$ ways
Problem 2
- how many no. of subsets of even size?
  - answer: $2^{n - 1}$ ways
    - 👨‍🏫 it doesn't apply for $n - 1 < 0$ (not natural number)
  - if $n = 6$
  - how many seq. have no 1s (= true)?
    - $1$ way: $\emptyset$
  - how many seq.. have exactly 2 1s?
    - i.e. different permutation of word: $000011$
    - $= \frac{6 !}{4 ! 2 !} = 15$
  - how many seq. have exactly 4 1s?
    - $= \frac{6 !}{2 ! 4 !} = 15$
    - (same combination, just 0s and 1s in worst change)
  - how many seq. have exactly 6 1s?
    - $1$ way: $[n]$
  - answer: sum of all
    - $1 + 15 + 15 + 1 = 32$
Problem 3
- how about for general $n$ ? (PA)
  - i.e. how many seq. of length $n$ have $2 i$ ones?
    - $2 i$ ones, $n - 2 i$ zeros
    - $= \frac{n !}{( 2 i )! ( n - 2 i )!}$
  - total sum: $\sum_{i = 0}^{n /2} \frac{n !}{( 2 i )! ( n - 2 i )!}$
    - 👨‍🏫 more complicated than thought :p
    - can we show that those two are same?
      - e.g. using induction, etc.
Problem 4
- for general $n$ (PM)
  - how many ways to assign fill in ${0, 1}$ to $n$ elements
    - for first $n - 1$ elements: we have freedom of choice
    - for the last element: we can only make $1$ choice
      - dependent on others
    - thus, $2^{n - 1}$

Practice Problems (Combination)

Problem 5
- prove $\sum_{i = 0}^{n} (i n) = 2^{n}$
  - 👨‍🏫 disclaimer: I don't expect you to be able to do any algebra
  - it's just different way of counting all subsets, so $2^{n}$
Problem 6
- prove $\sum_{i = 0}^{n} (- 1)^{i} (i n) = 0$
  - as there is a pair for each size
- 👨‍🏫 same number of even & odd subsets!
Problem 7
- prove $n (k n - 1) = (k + 1 n) (k + 1)$
- $n \geq k + 1$
- show it in combinatorial proof
  - recruiting captain & team, total of $k + 1$ people, from $n - 1$
  - choosing 1 captain from $n$ people & choosing the rest from $n - 1$ people
  - = choosing $k + 1$ people from $n$ people & choosing a captain from $n + 1$ people
Problem 8
- prove $(r n) = (r n - 1) + (r - 1 n - 1)$
- somewhat the same
  - no. of all teams
  - = no. of teams excluding a person A + no. of teams including a person A
Problem 9
- prove $\sum_{0 \leq i \leq j \leq n} (i n) (j i) = 3^{n}$
- RHS: 3 possibilities for a person
  - and all possible subsets
- LHS: for all group, where
  - $i$ people are in group $2, 3$
  - and $j$ people are in group $3$
- also: $\sum_{i = 0}^{n} (i n) \cdot 2^{i} = 3^{n}$
  - among $i$ people, there are total of $2^{n}$ ways to divide them between group $2, 3$
  - thus same
Problem 10
- how many solutions does
  - $x_{1} + x_{2} + x_{3} + x_{4} = 10$
  - have if $\forall i, x_{i} \in Z \land x_{i} \geq 0$ ?
- if $x_{1} = 10$ : 1 solution
- if $x_{1} + x_{2} = 10$ : $\forall x_{1}, \exists x_{2}$ w/ unique value
- what if $x_{1} + x_{2} + x_{3} = 10$ ?
  - 👨‍🏫 can we reduce the problem into permutation / combinations?
  - e.g. same as dividing $10$ balls with 2 bars
    - w/ 11 slots
    - or: aligning 10 balls and 2 bars
    - and the 10 balls & 2 bars are indistinguishable within them
    - $= \frac{12}{10 ! 2 !} = (2 12)$
- for general problem, $\sum_{i = 1}^{r} x_{i} = n$
  - where $x_{i} \geq 0$
  - there are $\frac{( n + r - 1 )!}{n ! ( r - 1 )!} = (r - 1 n + r - 1)$
  - as we have $r - 1$ bars
- or, the other way:
  - choosing $r - 1$ slots for bars from $n + r - 1$ slots
  - i.e. $(r - 1 n + r - 1)$
Problem 11
- prove $(r n) = (n - r n)$
- where $r \leq n$
- choosing $r$ people from $n$ people:
  - same as choosing $n - r$ people from $n$ whom you will exclude

COMP 2711H: Honors Discrete Mathematical Tools for Computer Science