Jus' Thinkin'
If order mattered, we would say select the first number (10 choices), then the second (9 choices) then the third (8 choices). So there would be 10 x 9 x 8 = 720 possible choices.
But because order does not matter, we have to take care of duplicates as you mentioned. How many duplicates are there for each set of three numbers? Well, again, we can choose one of the three as the "first", so there are 3 choices for that, then 2 choices for the "second" digit, then 1 choice for the last digit. There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits.
Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6.
720 / 6 = 120.
That's your answer.
These are what are mathematically called "combinations". You can use a formula involving factorials to determine the number of combinations.
In this case, we say this is "10 Choose 3" and write it as 10C3. That means from a set of ten (digits in
this case), choose 3 regardless of order.
The formula for nCm is
nCm = n! / [ m! x (n-m)!]
So in your question, we have
10C3 = 10! / [3! x (10-3)!] = 10! / [3! 7!]
= (10 x 9 x 8 x 7!) / [ (3 x 2 x 1) 7!]
= (720) / 6 ... because the 7! on the top and bottom cancels.
= 120 as we expected.
