To solve some mathematical problems, especially concerning in the probability theory, we have to compute the number of ways a set of objects can be arranged under some conditions. For this, we have to study different counting methods. The study of counting methods comes under the field of combinatorial algebra.

Let us consider three letters $A$, $B$ and $C$ and see in how many ways they can be arranged in a row. \[\begin{array}{c} ABC & BCA & CAB \\ ACB & BAC & CBA \end{array}\] Thus, the letters $A$, $B$, $C$ can be arranged in six ways.

This can be explained in a simple way. When we choose a first letter, there are three choices; we can choose any one of the three. For each choice of the first letter, there are only two letters left from which we have to choose the second letter. Hence, we can choose two letters in $3×2=6$ ways. Now, there is only one letter left for the third choice. Thus, these three letters can be arranged in $3×2×1=6$ ways.

This consideration leads us to the **basic principle of counting**.

## The Basic Principle of Counting states that:

If one thing can be done independently in $n_1$ different ways and if a second thing can be done in $n_2$ different ways and if a third thing can be done in $n_3$ different ways and so on (for any finite number of things), then the total number of ways in which all the things can be done in the stated order is \[n_1\cdot n_2\cdot n_3\cdot …\]

**Question: How many numbers between $4000$ and $5000$ can be formed with the digits $2,3,4,5,6,7$?**

Here, the numbers formed must be between $4000$ and $5000$, so the digit in the thousand’s place must be $4$. Thus, there is only $1$ way of filling the thousand’s place. After filling the thousand’s place, five digits are left. So, the hundred’s place can be filled up in $5$ ways. Now, there are only four digits left. So, the ten’s place can be filled up in $4$ ways and the unit’s place by $3$ ways.

Thus, numbers between $4000$ and $5000$ formed with the digits $2,3,4,5,6,7$ \[=1×5×4×3=60\]

## Factorial Notation

In many cases, we have to compute the products of the type $1\cdot 2\cdot 3$ or $1\cdot 2\cdot 3\cdot 4$ or $1\cdot 2\cdot 3\cdot 4\cdot 5$ etc. These products are denoted by a notation known as the **factorial notation**.

The product $1\cdot 2\cdot 3$ is denoted by $3!$ and is known as factorial $3$.

The product $1\cdot 2\cdot 3\cdot 4$ is denoted by $4!$ and is known as factorial $4$.

Similarly, the product $1\cdot 2\cdot 3\cdot 4\cdot 5$ is denoted by $5!$ and is known as factorial $5$.

Thus, if $n\in N$, then the continued product from $1$ to $n$ is known as factor $n$ and is written as $n!$.

Thus, \[\begin{array}{l} n! &= 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot…\cdot(n-1)\cdot n \\ 1! &= 1 \\ 2! &= 1\cdot 2 \\ 3! &= 1\cdot 2\cdot 3 \\ 4! &=1\cdot 2\cdot 3\cdot 4\end{array}\]

Also, we observe that \[\begin{array}{l} 5! &=1\cdot 2\cdot 3\cdot 4\cdot 5 \\ &= 5(1\cdot 2\cdot 3\cdot 4) \\ &= 5\cdot 4! \\ 11! &= 11\cdot(11-1)! \\ &= 11\cdot 10! \\ 25! &= 25\cdot(25-1)! \\ &= 25\cdot 24!\end{array}\] \[\therefore n! = n\cdot(n-1)!=n(n-1)\cdot(n-2)!\]

### Factorial $0$ or $0!$

By definition, $0!$ has no meaning. However, we can prove $0!=1$ mathematically as, \[\begin{array}{l} & 1! &= 1\cdot(1-1)! \\ \text{or,} & 1 &= 1\cdot 0! \\ \therefore & 0! &= 1\end{array}\]

**Next:** Permutation