**Treatment of similar items in permutations**

The difference between dealing with similar things and dissimilar things can be summarized by the following example. Suppose a bag contains three identical white balls except for their labels B1, B2 and B3. Now, we can clearly see that the balls can be arranged in 3! = 6 different ways namely B_{1}B_{2}B_{3}, B_{1}B_{3}B_{2}, B_{2}B_{1}B_{3}, B_{2}B_{3}B_{1}, B_{3}B_{1}B_{2} and B_{3}B_{2}B_{1}. When we remove labels from 2 balls, then 2 balls become similar in this case and hence the number of possible arrangements will be 3 (B_{1}BB, BB_{1}B, BBB_{1}). Here, we are neglecting the number of ways in which those 2 similar balls can be arranged and instead count them as 1. So, the net effect is that of dividing the total number of permutations by number of ways of arranging those 2 balls, which is 2!. Now, when we remove the labels from all the three balls, all three balls become identical and we can clearly see that there is only one way of arranging these balls as we would not be able to differentiate the arrangements listed above. Here, we are neglecting the number of ways in which the three similar balls can be arranged, which is 3!.

To generalize, when we are looking at arranging ‘n’ things, the number of possible arrangements will be n! when all things are dissimilar. When ‘p’ things are similar, then we need to divide the total possible arrangements by p! and hence the number of arrangements will be n!/p!. **Extending the logic, when ‘p’ things are of one kind and ‘q’ things are of the second kind with the rest of the things being distinct, then the number of different arrangements will be (n!/ (p! x q!)) and so on.**

### Treatment of repetition of items in permutations

The number of arrangements of n distinct items in r boxes where repetition of items is allowed can be figured out by reverting back to the original reasoning through which we derived the formula for permutations. If repetition is allowed, then the first box can be filled in ‘n’ ways and for each of these ‘n’ ways, the second box can be filled in ‘n’ ways and so on.

**Hence, if we have ‘r’ boxes in total, then the total number of ways in which the boxes can be filled using ‘n’**items,

**when repetition of items is allowed, will be n**.

^{r}