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Does anyone know how I can solve this problem? Any help would be great...... I cant seem to get my head around it.

As you know binary digits can only be either 1 or 0.

Say you had a 8 digit Binary number like a byte >>>>>> 0001 1000.

I'm trying to figure out an equation for the number of combinations you could get from an 8 digit binary number.

For example, if you had a two digit binary number, the binary combinations that you could have are:

00

01

10

11

Therefore the total combinations from a 2 digit binary number is 4.

Example 2

If you had a 3 digit number, the combinations would be:

000

001

010

100

101

111

110

011

Therefore the number of binary combinations from a 3 digit number is 8.

Example 3

If it were a 4 digit number, maximum binary combinations that you could have are either

0000

0001

0010

0100

1000

0111

0110

1111

1110

1101

1011

1001 Total maximum combination = 12

I Guess in a nutshell what im asking is .... if i had any number 6,7,15,8 or any number... how could i calculate the total maximum Binary combinations is there an equation to it ... I cant figure it out..ive tried for days now ;(

Chris Emerson
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13rianC
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2 Answers2

2

The number of numbers composed by d digits in base b is

b^d
Emanuele Paolini
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1

n - number of digits
b - base
^ - power

b^n

So your base is 2 (binary), and u want to check combinations for 8 digit number

2^8 = 256

apocalypse
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