Can someone explain Bios count or Boot count, something like that

Question

Like-- 0000=0 0001=1 00010=2

Answer 1

It is not Bios/Boot count, but "Binary count" - at least that is what the numbers you are showing are.

This simply refer to a "Binary" number system (base 2), as opposed to a "Decimal" (Base 10), or "Octal" (Base 8) or "Hexadecimal" (Base 16) uses.

We don't normally think of this, but we only agree on what the number 321 means because of convention i.e. that the base is 10. If the base were 5, 8 or whatever, that would be a totally different number.

For example: A number like 321 decimal really means:
3*10^2 + 2*10^1 + 1*10^0 = 300 + 20 + 1 = 321.
(where 3*10^2 means 3 multiplied by 10 squared)
Compare this to 321 octal (base 8):
3*8^2 + 2*8^1 + 1*8^0 = 192 + 16 + 1 = 209 decimal.
So 321 octal equals 209 decimal.

Of course we are all familiar to count in decimal:
|---------10-------|---..10..--|..
0,1,2,3,4,5,6,7,8,9,10,11,....19,20,..
To count in octal would be:
|---------8----|---..8..---|..
0,1,2,3,4,5,6,7,10,11,....17,20,..

So in decimal (base 10) the "carry" is after 10 numbers, in octal (base 8) the "carry" is after 8 numbers (so octal just require digits 0..7).

Likewise Binary is a base 2 system, and therefore the carry is after 2 numbers. This is exactly the same process as decimal/octal/hexidecimal/etc., so you have:
|2|-2--|---2----|
0,1,10,11,100,101,110,111
The only reason the numbers are acquiring more digits so quickly is because the base (2) is so low, but it still follows exactly the same "algorithm". Now 321 is not a valid binary number (only digits 0,1 can be used in a base 2 system), but lets use binary number 110101:
1x2^5 + 1x2^4 + 0x2^3 + 1x2^2 + 0x2^1 + 1x2^0
= 32 + 16 + 0 + 4 + 0 + 1 = 53 decimal.

So, to compare:
Hex/Dec/Oct/Bin
0 0 0 0
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 8 10 1000
9 9 11 1001
A 10 12 1010
B 11 13 1011
C 12 14 1100
...etc...

Answer 2

Thank you very very much!!!!!!!!!!!!!