Useful combinatorial identities (cont.)

• binomial theorem: let $n\in {\mathbb{Z}}^{+}$, then

  $$(x+y{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k{y}^{n-k}$$

  • $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$: often referred to as the binomial coefficient
  • combinatorial proof $$(x+y{)}^n=\underset{x^3+3x^{2}y+3x{y}^2+y^3}{\underbrace{\underset{x^2+xy+yx+y^2}{\underbrace{(x+y)(x+y)}}(x+y)}}\dots (x+1)$$
    • $k$ brackets contribute $x$, $n-k$ brackets for $y$

Problems

• how many subsets are there of a set consisting of $n$ elements?

  • intuition: there are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways of choosing $k$ elements from $n$
  • alternatively: you can assign following binary values to each elements
    • $b_i\{\begin{array}{l}1,\text{ if element i is included}\\ 0,\text{ if element i is not included}\end{array}$
    • and there are exactly $2^n$ combinations!

  ${\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^n$

• show that following equation holds true

  • ${\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})=0$
  • proof (algebraic) $0=(1-1{)}^n={\sum }_{k=0}^n(-1{)}^k\left(\genfrac{}{}{0ex}{}{n}{k}\right)$

• $\left(\genfrac{}{}{0ex}{}{n}{0}\right)+\left(\genfrac{}{}{0ex}{}{n}{2}\right)+\left(\genfrac{}{}{0ex}{}{n}{4}\right)\cdots =\left(\genfrac{}{}{0ex}{}{n}{1}\right)+\left(\genfrac{}{}{0ex}{}{n}{3}\right)+\left(\genfrac{}{}{0ex}{}{n}{5}\right)\dots$

  • can be shown from the previous proof
  • 👨‍🏫 can you show combinatorial proof?

Multinomial coefficients

• a set of $n$ distinct items: to be divided into $r$ different groups of respective size $(n_1,n_2,\dots ,n_r)$ where $\sum n_i=n$
  • how many divisions are possible? $$\begin{array}{rl}& =\left(\genfrac{}{}{0ex}{}{n}{n_1}\right)\left(\genfrac{}{}{0ex}{}{n-n_1}{n_2}\right)\left(\genfrac{}{}{0ex}{}{n-n_1-n_2}{n_3}\right)\dots \left(\genfrac{}{}{0ex}{}{n-n_1\cdots -n_{r-1}}{n_r}\right)\\ & =\frac{n!}{n_1!(n-n_1)!}\frac{(n-n_1)!}{(n-n_1-n_2)!n_2!}\dots \frac{n_r!}{0!n_r!}\\ & =\frac{n!}{n_1!n_2!\dots n_r!}\end{array}$$

• it's the same as arranging $n$ objects linearly ($n!$)
• and putting the first $n_1$ objects to group 1, next $n_2$ objects to group 2, etc.
  • as the order doesn't matter within the set, you can divide it by $n_1!,n_2!,$ and so forth

• and call it multinomial coefficients
• i.e. no. of ways in dividing $n$ elements into $r$ subgroups of each size $n_i$, total in $n$

Theorem

• multinomial theorem $$(x_1+x_2+\cdots +x_r{)}^n=\sum _{(n_1,\dots ,n_r):n_1+\cdots +n_r=r}(\genfrac{}{}{0ex}{}{n}{n_1,n_2,\dots ,n_r})x_1^{n_1}{x}_2^{n_2}\dots x_r^{n_r}$$
  • exercise: expand $(x_1+x_2+x_3{)}^5=?$
  • exercise 2: expand $(x_1+x_2+x_3+x_4+x_5{)}^3=?$

Problems

• police department: of 10 officers
  • each group: size of 5,3,2
  • how many divisions exist?
  • $\frac{10}{5,3,2}=\frac{10!}{5!3!2!}=7\ast 8\ast 9\ast 10/2=2520$
• ten children: divided into team A, B of 5 groups. how many divisions?
  • $\left(\genfrac{}{}{0ex}{}{10}{5,5}\right)=\frac{10!}{5!5!}=252$
• 10 children at playground divide themselves into teo teams of 5 each. how many divisions?
  • i.e. order of team A / B doesn't matter
  • $252/2=126$

Theorem

• theorem: there are $\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ distinct positive integer-valued vectors that satisfies the equation
  • $x_1+x_2+\cdots +x_r=n$
  • where $x>0$ for $i=1,\dots r$
  • proof: stars and bars method
    • put undistinguishable $n$ stars into $r$ bins (labeled from $\mathrm{1..}=r$)
      • s.t. $n$ bin is empty
    • let $x_i$ be the number of stars in the $i$th bin ($\mathrm{1..}=r$)
    • it's the same as placing $r-1$ bars on $n-1$ slots (between stars)
• theorem there are $\left(\genfrac{}{}{0ex}{}{n+r-1}{r-1}\right)$ distinct non-negative integer-valued vectors that satisfies the equation
  • proof

    • change "non-negative" into "positive" $$\begin{array}{rl}{x}_1+x_2+\cdots +x_r& =n\\ (y_1-1)+(y_2-1)+\cdots +(y_r-1)& =n\\ y_1+y_2+\cdots +y_r-r& =n\\ y_1+y_2+\cdots +y_r& =n+r\end{array}$$

    • and $y_i$ is a positive integer

    • thus $\left(\genfrac{}{}{0ex}{}{n+r-1}{r-1}\right)$

  • or, stars and bars method: now consider bars as objects too
    • it's basically aligning $n$ stars and $r-1$ bars

Problems

• investor: w/ 20K dollars to invest among 4 possible investment, in unit of thousand dollars.
  • if the total of 20k is to be invested: how man strategies (only holding) are possible?
    • 4 slots, with 20 stars
    • $\left(\genfrac{}{}{0ex}{}{20+4-1}{3}\right)=\frac{23!}{20!3!}=\frac{21\ast 22\ast 23}{6}=1771$
  • what if not all the money need be invested?
    • same as creating another investment of "nothing"
    • $\left(\genfrac{}{}{0ex}{}{20+5-1}{4}\right)=\frac{24!}{20!4!}=\frac{21\ast 22\ast 23\ast 24}{24}=10626$
• there are $n$ antennas with $m$ defective among them
  • how many orders there exist in which no two defectives are consecutive?
  • $n-m+1$ slots, $m$ bars
  • $\left(\genfrac{}{}{0ex}{}{n-m+1}{m}\right)$
• 👨‍🏫 other approach
  • .

Recap

• Ch. 1: used intuitive definitions to calculate probability
• translated "calculate probability" into counting no. of total outcomes & no. of interesting outcomes
• however: it can't deal w/ experiments where no. of total outcome & interesting outcomes are $\infty$
• we shall rigorously formalize probability
• terminologies
  • random experiments outcomes
  • event sample space
  • rigorous definition of probability

Terminology

• experiment: activity or procedure that produces distinct & well-defined possibilities
  • i.e. outcomes
  • basic object of probability
• sample space: set of ALL possible outcomes of an experiment
  • often denoted by $S$
  • e.g. head / tail on coin tossing
• event: any SUBSET $E$ of sample space
  • if random experiment produces an outcome in event $E$
    • we say "event $E$ occurs"
• size of sample space / event: can be finite or infinite