How many 3 - digit positive integers can be formed by using 1, 2, 3, 4, 5 without the repetition of digits?
Use the permutations formula:
5 nPr 3 = 60
5! / (5 - 3)! =120 / 2 = 60 such numbers
+773 use constructive counting and just have patience and try, that's how i do it.
+36916 5 choices for first digit leaves
4 choices for the second digit leaves
3 choices for the third digit....
5 x 4 x 3 = 60 possibilities
