## Arrangement allowing repetition

Given a set of $n$ *different* items, how may distinct arrangements of $n$ items are possible,
provided any item can used more than once?

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 items. So, we can do so in $n$ different ways. Now, note that the same item can be used more than once. Therefore, for filling the second vacant place, we can still use any of the $n$ items (including the item already used). So, we can fill the second vacant place again in $n$ ways.

Similarly, the third vacant place can also be filled in $n$ different ways, and so on. Finally, for the last vacant place we will again have $n$ different ways.

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. $n \times n \times n \times ... \times n$ ($n$ times) which
comes out to be $n^n$.

**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 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 allowed. We can fill the first place by any of the 4 given letters in 4 ways.
Since repetition is allowed, we can fill the second place again in 4 ways by any of the 4 different letters.
Similarly, the third place can be filled in 4 ways and finally, the last place can also be filled in 4 ways by any
of the 4 different letters.

Now, from the *multiplication theorem*, total number of distinct possible arrangements (with repetition)
is the product of the number of ways obtained in each step, i.e. $4 \times 4 \times 4 \times 4 = 4^4 = 256$.

(Answer: 256)