Question 2229
if the teacher wants all possibilities of 5 questions, even if repeated (just in different order) eg 12345 and 12354 are 2 different versions, then use permutations:

n=8, r=5, so we get {{{(n!)/(n-r)!}}}...8!/3! = 6720 different combinations of the 8 questions.

If however, the teacher wants all the combinations of 5 questions where 12345 and 12354 etc are considered the same, then there are 6720/5! = 56.

2a. odd digit first...4 possibilities,

so 4 then 7 then 7 then 7 then 7 possibilities with repetitions = 9604

2b. with repetitions not allowed....

we have 4 then 6 then 5 then 4 then 3 = 1440.

3. if number plate was to be odd, its last digit would have to be 1,3,5 or 7, so work that out .

Jon