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

**WASTAGE OF MEMORY. HOW? **

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**

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

**4**

**how?**

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

HH

HT

TH

TT

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-

00

01

10

11

**#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).**

**10^9**

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..**