From 1 to 20, you can pull out 15,504 different groups of 5 numbers,
and each group can be lined up in 120 different orders.
There are only five combinations: 1234, 1235, 1245, 1345 and 2345.
There are twelve possible solutions using the rule you stated.
There are 35C4 = 35*34*33*32/(4*3*2*1) = 52,360 combinations.
There are 840 4-digit combinations without repeating any digit in the combinations.
The number of combinations, denoted by 11C6 is 11!/[6!*(11-6)!] = 11*10*9*8*7/(5*4*3*2*1) = 462
There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
Including the leading zero, then the answer is 210. Or, allowing for repeating numbers, 10000.
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With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
First digit can be one of four, second can be one of the remaining three etc so 4 x 3 x 2 x 1 = 24 combinations. This is known as factorial 4 (or 4 factorial) and is symbolised as "4!".
The number of combinations you can make with the digits 1234567890 depends on how many digits you want to use and whether repetition is allowed. If you use all 10 digits without repetition, there are 10! (10 factorial) combinations, which equals 3,628,800. If you are choosing a specific number of digits (for example, 3), the number of combinations would be calculated using permutations or combinations based on the rules you set.
You can get only four combinations: They are: 11, 118, 119 and 1189. In a combination, the order of the digits does not matter.