3!(factorial) or six
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Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.
There are infinitely many possible answers. Here is one: 62 + 1
With two bits, there are (2^2) possible combinations, which equals 4. The combinations are: 00, 01, 10, and 11. Each bit can be either 0 or 1, leading to these four distinct configurations.
At least 11 of them.
-2