Well, honey, if you're looking for 7-number combinations out of 10 numbers, that's just basic math. You take 10 choose 7, which is 120. So, there you have it, 120 possible combinations. Math doesn't have to be complicated, darling.
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Without repeating any of the numbers (each used once) = 120 unique "sets" of 7 numbers
(i.e. 1234567, 1234568, 1234569, 1234560, 1234578, 1234579, 1234570, 1234589, 1234580, 1234590)
If the order matters, you have a much larger number of combinations. (see below) Each of the 120 sets can be arranged in 5040 ways.
For a set of N numbers, the possible combinations using K numbers is N! / K! x (N-K)!
10! = 3,628,800
7! = 5040
(10-7)! = 3! = 6
The shortcut is (10 x 9 x 8) / (3 x 2) = 720/6 = 120
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For rearrangement of numbers (ordered sets), there are 604,800 possible numbers
N! / (N-K) ! = 10! / 3! = 3,628,800 / 6 = 604,800
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For unlimited repetition, there are 10,000,000 (1 x 10^7)
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
You could make 10*10*10*26*26*26 combinations, or 17576000 combinations.
There are: 10C3 = 120
You will have 10 possibilities for each 4 spots (given that the position of the number matters), which gives you 10*10*10*10 = 10^4 = 10000 possible combinations. oh my god i was going to qrite them all down but nevermind!!!!!!!! that would take me forever