SOLUTION: 2. Consider the numbers 2, 3, 5, and 7. If repetition is not allowed, how many three-digit numbers can be formed such that; a. They are all odd? b. They are all even? c. They are

         Case (a)


Since they are all odd, it means that the last digit (in "units" position) is 3 or 5 or 7,
giving 3 options.

Then in "tens" position we can have 4-1 = 3 options for 3 remaining digits,

and in the "hundreds" position we can have 3-1 = 2 options for 2 remaining digits.


In all, it gives 3*3*2 = 18 possible numbers for case (a).   Thus case (a) is completed and answered.




         Case (b)


Since they are all even, it means that the last digit (in "units" position) is 2,
giving 1 option.

Then in "tens" position we can have 4-1 = 3 options for 3 remaining digits,

and in the "hundreds" position we can have 3-1 = 2 options for 2 remaining digits.


In all, it gives 1*3*2 = 6 possible numbers for case (b).   Thus case (b) is completed and answered.



         Case (c)


Since they are all greater than 500, it means that the first digit (in "hundreds" position) is 5 or 7,
giving 2 options.

Then in "tens" position we can have 4-1 = 3 options for 3 remaining digits,

and in the "ones" position we can have 3-1 = 2 options for 2 remaining digits.


In all, it gives 2*3*2 = 12 possible numbers for case (c).   Thus case (c) is completed and answered.