License Plate ProbabilityA certain State has license plaes showing three numbers and three letters. How many possible combinations are available?

#Without Replacement 
numbs<- choose(10, 3)
letts <- choose(26, 3)
total <- numbs * letts

# With Replacement
numbs_rep <- 10^3
letts_rep <- 26^3
total_rep <- numbs_rep * letts_rep

# Random order of numbers and letters 
#
total_ro_wo <- factorial(36)/ (factorial(3)^2)

a.) If numbers must come before letters

There are ten numbers and 26 letters, you must pick sets of three from all possible combinations within each and then then all possible combinations of the triplets

Without replacement

Number Sets: r numbs

Letter Sets: letts

Answer \(NumberSets \times LetterSets =\) 312000

With replacement

Number Sets: r numbs_rep

Letter Sets: letts_rep

Answer \(NumberSets \times LetterSets =\) 17576000

b.)

Were the number of choices for each position is 36 and the number for each type of character is 3 then the number of possible plate numbers in random distribution wiht three of each is the number of combinations of plates you can make with 36 characters, limited by the constraints of the number of 3 number and 3 letter combinations. This relies on no replacement

Answer is:
\(number_Choices!/(numbers! \times letters!) = 36!/ (3!3!)=\) 10333147966386144222209170348167175077888