Arrangement allowing repetition

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

For solving this question, let us suppose that there are nn vacant places placed one after another. Now, we can fill the first vacant place by choosing any of the given nn different items. So, we can do so in nn 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 nn items (including the item already used). So, we can fill the second vacant place again in nn ways.

Similarly, the third vacant place can also be filled in nn different ways, and so on. Finally, for the last vacant place we will again have nn 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×n×n×...×nn \times n \times n \times ... \times n (nn times) which comes out to be nnn^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. 4×4×4×4=44=2564 \times 4 \times 4 \times 4 = 4^4 = 256.
(Answer: 256)


Back to Main index

Go to Home