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.

  1. 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.

  1. 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

  1. 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

  1. 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

  1. 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

Coins, Dice and Cards

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