If a combination lock has 3 numbers slots ranging from 0 - 9 how many different combinations are there ?

Answer 1

0-9 is ten digits and there are three of them. So, the total number of permutations (that is what they are), is 1000. From 000 to 999.

Answer 2

1000,


if you count 0 , if 0 is on the dial then yes.


10^3 or if its a total of nine numbers 9^3(= 729)

Answer 3

Let
S1 = 1st slot
S2 = 2nd slot
S3 = 3rd slot

combination = S1 S2 S3

Assuming the combination lock with the knob where you
can turn clock-wise or counter-clockwise

there are 20 choices for S1 (0-9, clock-wise and counter-clock-wise)
20 choices for S2
20 choices for S3

total # of choices is
20 * 20 * 20 = 8000

Answer 4

i got 1000 here's how i solved it...it's kinda just logic so the first cylinder has 10 numbers, so do the second and third okay. so the first cylinder has 0,1,2,3,4,5,6,7,8,9 that is ten digits. the second cylinder has ten digits as well. thus, each of the ten digits in the first cylinder can be paired to any of the ten digits in the second cylinder. the 0 on the first cylinder can be paired to all ten digits of the second cylinder. same goes with all the rest of the digits on the first cylinder. so if you have 10 digits, and 10 possible combinations per digit, you have 10 digits x 10 combinations each=100 combinations so you have 100 possible combinations for the first 2 cylinders. the third cylinder has 10 digits as well. each of these 10 digits can be combined with each of the 100 combinations of the first and second cylinder. therefore you have 10 digits in the third cylinder x 100 combinations you can combine the digits with = 1000 combinations :D i hope i got to explain it well...

Answer 5

10 * 10 * 10 = 1000

Answer 6

asume numbers are repeated
10 x 10 x 10 = 1000

note: including 000

Answer 7

9x9=81x9=729