How many 3 number cominations are there?

Answer 1

Depends on the range of numbers (0-9) you would like to choose from and depends on any restrictions you would want. For example:

1. Range: 0-9, any number can be repeated
possible 3-digit combinations = 1000 (10x10x10)

2. Range: 0-9, no number can be repeated
possible 3-digit combinations = 720 (10x9x8)

Answer 2

If not including zero as first number then
9 X 10 X 10 = 900
1 THROUGH 99 DON'T COUNT. 99
1000 DON'T COUNT 1
So then 900 numbers. 1000 - 100 = 900

Another way to explain this is...

9 for the nine options of numbers for the first # (1-9)
10 for the ten options of number for the second # (0-9)
10 for the ten options of numbers fo the third # (0-9)

OR

If zero included for first number then 10 X10 X10 = 1,000
Examples: 000, 012, 098, 036, 047.

If you are woundering.. "there are only 999 numbers because 1,000 isn't 3 digit" you're right but if zero could be a first number then 000 would count.

Answer 3

# of possible term1 x # of possible term2 x # of possible term3

for each term you said that there were 10 digits so
10 x 10 x 10 is correct

If however the numbers had to be 1 or 0 (binary code)
2 choices for each term
2x2x2=8

Answer 4

There are a thousand combinations of three digits. (Each can range from zero to nine, and can repeat)

If you have a set of n items, and take n of them at time, then there is exactly one combination, since order is not important.

What I am trying to say is that you haven't given enough information to solve this problem.