+0  
 
+1
55
4
avatar

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

 Jan 3, 2020
 #1
avatar
0

Use constructive counting.

 Jan 3, 2020
 #2
avatar
0

Use the permutations formula:

 

5 nPr 3 = 60

 

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

 Jan 3, 2020
 #3
avatar+203 
+1

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

 Jan 3, 2020
 #4
avatar+20217 
+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

 Jan 3, 2020

18 Online Users

avatar
avatar