There are (2^5) bit strings of length 5, as each bit can be either 0 or 1. Therefore, the total number of bit strings is (32).
Three of them ... the binary numbers 3, 5, and 6. 011 101 110
8. Algebraic way of solving: 1 * 2 * 2 * 2 * 1 Long way of solving: 10000 10010 10110 11110 10100 11100 11000 11010
To find the number of strings of length 5 generated by the productions S → a0b, a → bb0, and b → aa1, we can analyze the derivations. Starting with S, we can replace it with a0b or explore the substitutions for 'a' and 'b'. Each substitution increases the length of the string, so we need to systematically track how the productions can be combined to form strings of exactly 5 characters. After evaluating the combinations, we find that the total number of strings of length 5 is 8.
Five (5) have one or the other but not both. Six (6) have both. Total of eleven (11).
Three of the bits are already determined. That leaves 5 .25 = 32
5
4? 5?
5
The longest factor string for 45 is 3 x 3 x 5
the total 128 ^5 the strings without @ at all 127^5 to get the strings that has at least @ once 128^5 - 127^5
Yes, many have 5 strings
Actually, lengths of a bass does not depend on the number of strings. Basses can be of different length but this has no connection with the number of strings.