Question

1. A character in a ASCII system has size 1 byte or 8 bit (1 or 0)

A). How many bit patterns are formed?

B). How many bit patterns have 3 bits 1?

C). How many bit patterns have an even number of bits 1?

Answer 1

Solution :

A) Number of bit patterns that can be formed : 2⁸ bit patterns can be formed using 8 bits.

With n bits, 2ⁿ bit patterns are possible. So with 8 bits 2⁸ bit patterns possible.

B) Number of bit patterns that have 3 bits 1 :

out of 8 bits , any 3 bits are selected to be 1 and other bits has to be 0.

There are $\binom{8}{3}$ ways to select 3 bits out of 8 bits. (we are selecting 3 bits to be 1 out of 8 bits. Which means rest 5 bits are 0 and we choose 3 bits to be 1).

$\left [ \binom{n}{r} =\frac{n!}{\left ( n-r \right )! *r!} \right ]$

hence there are $\binom{8}{3}$ bit patterns possible which has exactly 3 bits 1.

C) Bit patterns with even number of bits 1 :

even number of bits 1 = 0 number of bits 1 + 2 number of bits 1 + 4 number of bits 1 + 8 number of bits 1

$=\binom{8}{0} +\binom{8}{2} +\binom{8}{4}+\binom{8}{8}$

= 1+ 28 + 70 + 1

= 100

hence there are 100 bit patterns that have an even number of bits 1.

if you have any doubts, you can ask in comment section.