## Arrangement without repetition

Given a set of $n$ different objects, how may distinct arrangements of the objects are there?

For solving this question, let us suppose that there are $n$ vacant places placed one after another. Now, we can fill the first vacant place by choosing any of the given $n$ different objects. So, we can do so in $n$ different ways. Now, for filling the second vacant place, we cannot use the one object which we have already used because repetition is not allowed. So, we can fill the second vacant place in $(n -1)$ ways.

Similarly, the third vacant place can be filled in $(n - 3)$ different ways, and so on. Finally, for the last vacant place we will have only $1$ way since all the other $n -1$ objects are already used for the previous places and we are left with only one object.

Now, from the multiplication theorem, total number of distinct possible arrangements is the product of the number of ways obtained in each step, i.e. $n \times (n - 1) \times (n - 2) \times ... \times 1$.
We will see this product a lot in this tutorial, so much so that, in mathematics there is a separate notation for it, written as $n!$ (read as "$n$ factorial").

Example:
Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word LABS, where the repetition of the letters is not allowed.

Soln. There are as many words as there are ways of filling in 4 vacant places ( _ _ _ _ ) by the 4 letters (L,A,B,S) , keeping in mind that the repetition is not allowed. We can fill the first place by any of the 4 given letters in 4 ways. Once we do that, only 3 letters are left now and therefore, we can fill the second place in only 3 ways. Similarly, the third place can be filled in 2 ways and finally, the last place can be filled in only 1 way by the last remaining letter.

Now, from the multiplication theorem, total number of distinct possible arrangements is the product of the number of ways obtained in each step, i.e. $4 \times 3 \times 2 \times 1 = 24$.