Best Answer

Five (5) have one or the other but not both.

Six (6) have both.

Total of eleven (11).

Q: How many bit strings of length 8 contain either three consecutive 0s or four consecutive 1s?

Write your answer...

Submit

Still have questions?

Continue Learning about Other Math

Length and width.

there are 128 (2 to the power of 7) bit strings of length 7

-- 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.

2 x 2 x 13

Length = 4, Width = 2

Related questions

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.

56 The number of triples of 1s on 8 bits

Character string values storage:1. CHAR:§ Stores strings of fixed length.§ The length parameter s specifies the length of the strings.§ If the string has smaller length it padded with space at the end§ It will waste of a lot of disk space.§ If the string has bigger length it truncated to the scale number of the string.2. VARCHAR:§ Stores strings of variable length.§ The length parameter specifies the maximum length of the strings§ It stores up to 2000 bytes of characters§ It will occupy space for NULL values§ The total length for strings is defined when database was created.3. VARCHAR(2):§ Stores strings of variable length.§ The length parameter specifies the maximum length of the strings§ It stores up to 4000 bytes of characters§ It will not occupy space for NULL values§ The total length of strings is defined when strings are given

2X26

52

Length and width.

1024

there are 128 (2 to the power of 7) bit strings of length 7

By the sum rule we can count the number of strings of length 4 or less by counting the number of strings of length i, for 0 <= i <= 4, and then adding the results. Now there are 26 letters to choose from, and a string of length i is specified by choosing its characters, one after another. Therefore, by the product rule there are 26^i strings of length i. The answer to the question is thus: sum i=0 to 4 ( 26^i = 1 + 26 + 676 + 17576 + 456976) = 475,255.

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.

210=1024

The law of vibrating strings is the vibrational mode of a string that is stretched. The wavelength is twice the length of the string.