If n is even then number of palindromes is 2^(n/2).
If n is odd then number of palindromes is 2*2^[(n-1)/2].
The number of different bit strings of length 7 can be calculated using the formula 2^n, where n is the length of the bit string. In this case, with a length of 7, there are 2^7 = 128 different bit strings possible. This is because each bit in the string can have 2 possible values (0 or 1), and there are 7 bits in total.
-- There are 256 bit strings of length 8 . -- There are 4 bit strings of length 2, and you've restricted 2 of the 8 bits to 1 of those 4 . -- So you've restricted the whole byte to 1/4 of its possible values = 64 of them.
You are essentially asking how many different bit strings of length 9 are there, as the first 3 bits are fixed. The answer is 2^9 = 512.
n+1 (counting the empty string)
That's just the highest number that can be counted with 6 bits ... 127 .
Every bit can either be a 0 or 1. So to find the amount of bit strings of length either, you do 2length to find the amount of bit strings there are of a given length.
1024
210=1024
The number of different bit strings of length 7 can be calculated using the formula 2^n, where n is the length of the bit string. In this case, with a length of 7, there are 2^7 = 128 different bit strings possible. This is because each bit in the string can have 2 possible values (0 or 1), and there are 7 bits in total.
Half of them.
There are no zero-length strings that start with 1 bit or end with 2 bits. In a zero-length string, there are no bits at all.
-- There are 256 bit strings of length 8 . -- There are 4 bit strings of length 2, and you've restricted 2 of the 8 bits to 1 of those 4 . -- So you've restricted the whole byte to 1/4 of its possible values = 64 of them.
There are 210.
1024 of them.
You are essentially asking how many different bit strings of length 9 are there, as the first 3 bits are fixed. The answer is 2^9 = 512.
n+1 (counting the empty string)
its 2 raised to power 8 i.e 256