These problems are to enhance the theoretical learning. Solutions are provided under the questions. There might be R codes to replicate what is done on paper.

Suppose I toss a coin, roll a die and draw a card from the deck. How many different number of outcomes are there for this experiment?

Solution: Multiplication rule. \(n_1n_2n_3 = 2.6.52 = 624\).

`n1 <- 2 #A coin toss has two potential outcomes. n2 <- 6 #A die roll has six potential outcomes. n3 <- 52 #A card draw has 52 potential outcomes.`

In how many ways can I order the Teletubbies? (Tinky-Winky, Dipsy, Laa Laa and Po) For instance, (TW - Dipsy - Po - Laa Laa) is an ordering and (Dipsy - Po - TW - Laa Laa) is another.

Solution: Permutation rule. \(n! = 4! = 24\)

`n_tubbies <- 4 #Number of teletubbies factorial(n_tubbies) #By permtuation it is 4!`

`## [1] 24`

I want to reorder the letters of the phrase “GOODGRADES”. In how many ways can I do it?.

Solution: Remember the permutation rule with identical items. There are two “G”s, two “D”s and two “O”s. Remember the formula \(\dfrac{n!}{n_1!n_2!\dots n_k!}\). So the result should be \(\dfrac{10!}{2!2!2!1!1!1!1!} = 453600\).

`the_phrase <- "GOODGRADES" freq_table <- table(strsplit(the_phrase, split = "")[[1]]) #Let's create a frequency table first print(freq_table) #Let's show it`

`## ## A D E G O R S ## 1 2 1 2 2 1 1`

`the_dividend <- factorial(nchar(the_phrase)) #Dividend part is 10 characters so 10! the_divisor <- prod(factorial(freq_table)) #Get multiplication of factorials for the divisor the_dividend/the_divisor`

`## [1] 453600`

I want to make two letter words from “GRADES” such as “GA”, “ED” or “DE” (it doesn’t have to make sense). Find the number of permutations.

Solution: Permutation of \(r\) items from \(n\) items is \(\dfrac{n!}{(n-r)!}\). So the result is \(\dfrac{6!}{4!} = 30\).

`the_phrase<-"GRADES" letter_length <- 2 #We want two letter words #Since all letters are different no need for special permutation. factorial(nchar(the_phrase))/factorial(nchar(the_phrase)-letter_length)`

`## [1] 30`

Suppose I am drawing a hand of 5 cards from a playing deck of 52 cards. How many different hands there can be? (Each card is different. See the bottom of this document for details.)

Solution: Since in a hand you do not care for the order, it is the combination \(\binom{52}{5} = \dfrac{52!}{(52-5)!5!} = 2598960\).

`#Combination (a.k.a binomial coefficient) function is choose choose(52,5)`

`## [1] 2598960`

When questions mention about coins, dice and cards they are commonly referred items. Nevertheless, you can refer to .

- Coin tosses: Two possible outcomes. Heads or Tails.
- Dice rolling: Six possible outcomes. 1-2-3-4-5-6.
- Card drawing: 52 possible outcomes. There are 4 types (clubs, diamonds, spades and hearts) and 13 ranks for each type. (A)ce - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - (J)ack - (Q)ueen - (K)ing.