Almost every book on computer applications at the school level skip the beauty of the mathematics behind it

In fact basic counting principles are very easy to understand & should be part of every curricula

Excluding it leaves the interesting & easy but important basic part

The concept of 2

Most of the computational data is stored & evaluated as binary numbers

Each digit of a binary number(base 2) is either 0 or 1(one less than the base of the system)

Why just 2 ?

A computer or any other electronics device generally uses logically 0 & 1 to represent either if any 2 states-one for 0; the other for 1

It is not necessary that 0 always represents low/off/false. That depends on the predefined logic

Most common situations are such that there are only two possible answers/outcomes/values for it

For example, a CFL can either glow(ON) or not(OFF) depending on the switch position

A coin generally when tossed either gives a head or a tail; in many movies we have seen a funny situation arising when the coin lands on its rim(neither head nor tail)

If we also consider this outcome, how many possibilities are there? Simply 3

Taking the base of the number system to be equal to the number of possibilities we reduce the chances of redundant data & wastage of memory


Here I will go into very basic counting problems to explain the memory allocation. It is part of the +2 mathematics syllabus as well

#1. A coin on being tossed either results in a head or a tail. How many outcomes are possible?


#2. Two coins are tossed. A coin on being tossed either results in a head or a tail. How many outcomes are possible?



Representing head by H & tail by T, I point out the possibilities





Now you must be thinking that this is such a lengthy process ! In face there is a theorem in mathematics that states that no matter how many independent processes are carried one after the other the outcome is their product.

Now the process of tossing a coin doesn’t affect the results we obtain on tossing the other coin; hence they are independent processes.

#3. A two digit binary storage unit can store how many numbers?

4; this is a situation analogous to #2

the possibilities are-





#4. A 100 digit binary storage unit can store how many numbers?

2^100, its a huge number

quite boring. I too feel the same. Answer the next question & you will not feel the same.

#5. How many mobile numbers are possible in India at present? Assume that the first digit is always 9, the rest can be anything(0 to 9).


Recently it was reported in the news papers that TRAI is considering adding an extra 9 to the beginning

of all mobile numbers. Any idea why??

In our case it was a ten digit mobile number beginning with 9, now its a eleven digit mobile number beginning with 9. So how many mobile numbers are possible now? 10^10; 10 times the previous

NOTE : you may ask why not simply let the first digit start with any number(0-9). Then there can be 10^10 number of possible 10-digit mobile numbers. True

I don’t know the reason & will update this part of the text after finding it out

#6. Now this final question will illustrate the memory wastage on using say, a base-7 number system, to store values that have 2 possible outcomes.

How many possible values can a 10-digit base 7 number have? This is exactly its storage capacity.

Obviously, 7*7*7…..=7^10

In how many possible ways, can a combination of a serial-array of ten bulbs be lighted? This is exactly the storage that will be used.

Ten bulbs A, B, C ,D, E, F, G, H, I, J are arranged in a line ABCDEFGHIJ. The various ways they can be lighted equals 2*2*2…=2^10, each bulb can either be on or off.

So if the base-7 system is used to store values then the wasted memory=?

7^10 – 2^10, a huge number.

#7. then why are hexadecimal & octal numbers used by computers?

This is the basic concept, that once understood , will not ever create confusion.

The last question, asked for a storage of a 10-digit binary linear array into a 7-digit octal linear array.

However, a binary number can be converted to any base. Not just binary numbers, any number in a given base-n can be converted to its corresponding equal number in another base-m.

What is the binary equivalent of the decimal number 10? 1010

how? It can be found in any standard textbook

10=1*(2^3) + 0*(2^2)+ 1*(2^1)+0*(2^0)

noticed any pattern? Each digit of the decimal number is multiplied by 2 raised to its positional-index(numbered from 0 & not 1).

noticed anything else?

The same number required 4-digits in the binary system but required only 2 in the decimal system.

In the hexadecimal system, the number of digits would be even less-one for the decimal number 10, represented by the hexadecimal number A.

So what exactly is our benefit if the number of digits s less or more?

Less number of digits implies less number of electronics signals have to be transmitted

next-cylinders, storage units(mb,gb,etc), meta data, why does a 250 GB hard disk have less available space, etc..