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.

2. 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

3. 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

4. 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

5. 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