# How many bit strings of length eight either start with a 1 bit or end with the two bits 00?

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Number of bit strings of length 8 that start with 1 and end with 00: 25 = 32. Applying the subtraction rule, the number is 128 + 64 − 32 = 160.

Also to know is, how many bit strings of length ten either start with a 1 bit or end with the two bits 00?

Bits are either 1 or 0, so there are 2 choices per bit, and 8 bits to choose. Therefore, there are 28 = 256 bit strings of length ten that begin and end with a 1.

Similarly, how many bit strings of length 8 or less are there? There are 28 which is 256. That means there are 256 different values you can store in a byte, since a byte is eight bits. There are 256 eight-bit ascii codes, for instance.

Considering this, how many bit strings of length 8 begin and end with a 1?

My answer would be: 26. Because we know, that the bit starts with 1 and ends with 1: Which means, that 2 of the length 8 is used, also there are 6 positions remaining, which is 26=64.

How many 8 bit strings have at least two consecutive 0's or two consecutive 1's?

There are only two 8-bit strings in which the bits alternate: 10101010 and 01010101. The total number of 8-bit strings with no restrictions is 2^8. Therefore the number of 8-bit strings which have at least two consecutive 0's or at least two consecutive 1's is 2^8 - 2.

### How many bit strings of length 10 begin and end with a 1?

We are asked about 10 bits, but the first and the last bits are already chosen for us. Bits are either 1 or 0, so there are 2 choices per bit, and 8 bits to choose. Therefore, there are 28 = 256 bit strings of length ten that begin and end with a 1.

### How many bit strings of length 9 are there?

a) There 2*2*2*2*2*2 9 times = 2^9 = 512 bit strings of length 9.

### How many bit strings of length n are there?

There are two bit strings of length 1, '0' and '1'. In general, there are 2^n bit strings of length n.

### How many 5 bit strings are there?

We have 5 bits, and each can either be a 0 or a 1. So there are 2 choices for the first bit, 2 choices for the second, and so on. By the multiplicative principle, there are [Math Processing Error] 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 5 = 32 such strings. Finding the number of 5-bit strings of weight 3 is harder.

### How many strings are there of lowercase letters of length four or less not counting the empty string?

How many strings are there of lowercase letters of length four or less, not counting the empty string? The objective is to find the number of strings that are of lowercase letters of length four or less excluding the empty string. The number of possible letters in English alphabets is 26.

21

### How many 8 bit numbers are odd or have exactly 4 bits that equal 0?

The answer is there are 70 eight-bit binary integers of all the possible 8-bit binary integers which have exactly four zeros. Damereau-Levenshtein distance is used to calculate the minimum number of operations to transform one string.

### How many 4 bit binary strings are there?

By dividing a binary number up into groups of 4 bits, each group or set of 4 digits can now have a possible value of between “0000” (0) and “1111” ( 8+4+2+1 = 15 ) giving a total of 16 different number combinations from 0 to 15.

Decimal Number 4-bit Binary Number Hexadecimal Number
5 0101 5
6 0110 6
7 0111 7
8 1000 8

### What is bit string in discrete mathematics?

A bit string is a sequence of bits. Bit strings can be used to represent sets or to manipulate binary data. The elements of a bit string are numbered from zero up to the number of bits in the string less one, in right to left order, (the rightmost bit is numbered zero).

### How many six bit strings are there that begin with 01?

Similarly, there are 26 bit strings that end with 01. The sum 27+26 double-counts the bit strings that start with 1 and end with 01. There are 25 of these, so there are 27+26−25 bit strings that start with 1 or end with 01.

### How many strings of five ascii characters contain the character at sign at least once Note there are 128 different ascii characters?

There are 5 ways to place this character, because the string has a length of 5. The remaining characters may or may not be an @ symbol. Each of the four remaining characters can be chosen in 128 different ways. By the rule of product, there are 5 * 128 * 128 * 128 * 128 = 5*128^4 such strings.

### How many different passwords are there that contain only digits and lower case letters and satisfy the given restrictions a length is 6 and the password must contain at least one digit b length is 6 and the password must contain at least one digit and at least one?

How many different passwords are there that contain only digits and lower-case letters and satisfy the given restrictions? Length is 6 and the password must contain at least one digit. The total number of passwords with no restrictions is 36^6. The number of passwords with no digits (i.e., only letters) is 26^6.