How many 3 - digit positive integers can be formed by using 1, 2, 3, 4, 5 without the repetition of digits?

Guest Jan 3, 2020

#2**0 **

Use the permutations formula:

5 nPr 3 = 60

5! / (5 - 3)! =120 / 2 = 60 such numbers

Guest Jan 3, 2020

#3**+1 **

use constructive counting and just have patience and try, that's how i do it.

whymenotsmart Jan 3, 2020

#4**+1 **

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

ElectricPavlov Jan 3, 2020