Answer to Question #282376 in Discrete Mathematics for Neha

Question

bit is either 0 or 1: a byte is a sequence of 8 bits. Find

(a) the number of bytes that can be formed

(b) the number of bytes that begin with 11 and end with 11

(c) the number of bytes that begin with 11 and do not end with 11 and

(d) the number of bytes that begin with 11 or end with 11.

Answer 1

(a) Select a digit from two digits (0 or 1) for the first place, a digit from two digits (0 or 1) for the second place, a digit from two digits (0 or 1) for the third place, and so forth.

The number of possible different bytes is

$$2^8=256$$

(b) There are 4 vacant positions.

The number of bytes that begin with 11 and end with 11 is

$$2^4=16$$

(c) The number of bytes that begin with 11 is

$$2^6=64$$

The number of bytes that begin with 11 and end with 11 is

$$2^4=16$$

The number of bytes that begin with 11 and do not end with 11 is

$$2^6-2^4=64-16=48$$

(d) The number of bytes that begin with 11 is

$$2^6=64$$

The number of bytes that end with 11 is

$$2^6=64$$

The number of bytes that begin with 11 and end with 11 is

$$2^4=16$$

The number of bytes that begin with 11 or end with 11 is

$$2^6+2^6-2^4=64+64-16=112$$