Factorial of a number n is product of all positive non-zero numbers less than or equal to $n$. The factorial of $n$ is denoted by $n!$. The $(!)$ symbol is used to denote a factorial. Factorial of any number can be represented as product of the number and factorial of a number less than it by one. It can be written as,
$n!$ = $n\ \times\ (n -1)!$
Example: Find the value of $(\frac{3}{2})!$

Hence, $(\frac{3}{2})!$ = $\frac{3}{2}$ $\times$ $(\frac{1}{2})!$

Now, $(\frac{1}{2})!$ = $\frac{1}{2}$ $\sqrt{\pi}$

Then we have,

$(\frac{3}{2})!$ = $\frac{3}{2}$ $\times$ $\frac{1}{2}$ $\sqrt{\pi}$ = $\frac{3}{4}$ $\sqrt{\pi}$

Word Problems

Problem 1: 

A game has to be made from marbles of three colors, yellow, blue and green where three marbles has to be kept one upon another. In how many ways these marbles can be arranged?


There are three colors, and hence the marbles can be arranged in $3!$ ways.

Marble can be arranged in $3!$ = $3.2.1$ = $6$ ways.
Problem 2: 

Four guys have to stand in a queue to get football match tickets. In how many ways they can form the queue?


The number of ways the queue is formed is $4!$.

Hence, number of ways queue is formed = $4!$ = $$ = $24$
Problem 3: 

In how many ways the letters of the word "ALPHA" can be rearranged?


Total number of letters = $5$

Number of repeated letters = $2$, that is, $A$ is repeated twice

Number of ways the letters can be rearranged = $\frac{5!}{2!}$ = $\frac{}{2.1}$ = $60$

Hence, the letters can be rearranged in $60$ ways.
Problem 4: 

Four people $A,\ B,\ C$ and $D$ are sitting in a circular arrangement. In how many ways their seating can be arranged?


For $n$ different things, the number of circular arrangement is given by $(n - 1)!$.

Hence, total number of arrangements = $(4 - 1)!$ = $3!$ = $3.2.1$ = $6$