24
6, as long as they are all different
None. 3 digit numbers are not divisible by 19 digit numbers.
There are 5 numbers which can make the 3 digit numbers in this example. Therefore each digit in the 3 digit number has 5 choices of which number can be placed there. Therefore number of 3 digit numbers = 5 x 5 x 5 = 125
106------------------------------------------------------------------------------------------------From Rafaelrz.You can make, 4P3 = 4!/(4-3)! = 24, different 3 digit numbers using 1, 2, 3 and 4.
24
You can make 4*3*2/2 = 12 numbers.
6, as long as they are all different
None. 3 digit numbers are not divisible by 19 digit numbers.
There are 5 numbers which can make the 3 digit numbers in this example. Therefore each digit in the 3 digit number has 5 choices of which number can be placed there. Therefore number of 3 digit numbers = 5 x 5 x 5 = 125
106------------------------------------------------------------------------------------------------From Rafaelrz.You can make, 4P3 = 4!/(4-3)! = 24, different 3 digit numbers using 1, 2, 3 and 4.
12,13,14 21,23,24 31,32,34 43,42,41 12 two-digit numbers
24 = 4*3*2*1
There are 151 3-digit numbers that are divisible by 6.
If you want 4-digit numbers, there are 24 of them.
4C3 = (4!)/(4 -3)!3!] = (4 x 3!)/(1!3!) = 4 ways
9