how many input combinations can be there in the truth table of a logic system having (n) input binary values
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You can arrange a n-bit number in two raised to the n different possibilities : 2^n. Let's suppose a 3-bit number and write all its possibilities.
000, 001, 010, 011, 100, 101, 110, 111
These are 8 possibilities = 2*2*2 = 2^3
This also can be seen as the permutation with repetition of three events over two possibilities. These events are the total number of digits, and each digit, since it is binary, has two possibilities, so:
2 possibilities of first digit * 2 possibilities of second digit * 2 possibilities of third digit = 8 total possibilities = 2^3
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