Permutations of Identical Objects (Part 2)


Till now, we had identical objects of only one type. That is, only the red balls were identical, and the rest were all different.

This lesson will consider a couple of more cases, which involve identical objects of more than one type. For example, 3 red balls and 2 green balls. Or 2 red, 3 green and 4 other different colored balls.

Suppose we have to arrange 5 balls in a row of which 3 are red and the other 2 are green. How many different arrangements are possible?

We’ll designate 5 spaces as before, each of which is to be filled by one ball.

First, we’ll place the red balls in any 3 of the 5 spaces. This can be done in 5C3 ways. Here’s one of them.

Red Red Red

Next, in the remaining 2 boxes, we’ll place the remaining 2 balls. This can be done in only 1 way (as both of them are identical)

Red Red Green Red Green

 

And that’s it. The total number of arrangements will be 5C3 x 1 or \( \frac{5!}{3!2!} \)

We could also have places the green balls first, in 5C2 ways (select 2 out the 5 spaces and put one ball in each), and then place the red balls in the remaining spaces. The number of arrangements remains the same: 5C2 or \( \frac{5!}{2!3!} \)

Now let’s take another case: 7 balls of which 3 are red, 2 are green and the remaining are of different colors.

We begin by placing the red balls, in 7C3 ways.

Red Red Red

Then place the green ones, in 4C2 ways. (As we have 4 boxes remaining of which have to select 2)

Red Red Red Green Green

Finally, we’ll place the remaining distinct balls. This time we have 2! ways, because the arrangements of these 2 balls matter.

Red Red Pink Red Green Blue Green
Or
Red Red Blue Red Green Pink Green

 

The total number of arrangements are 7C3 x 4C2 x 2! = \( \frac{7!}{3!4!} \times \frac{4!}{2!2!} \times 2! = \frac{7!}{3!2!} \)

Do you see a pattern? In the first case we had 5 balls, of which 3 were red and 2 were green. The number of arrangements came out to be \( \frac{5!}{3!2!} \)

Had all of the 5 balls been distinct, our answer would have been 5!. Now if 3 of them were replaced with identical red balls, the answer changed to 5!/3!, and if the remaining 2 were also replaced with identical green balls, the number of arrangements changed to \( \frac{5!}{3!2!} \)

In the second case we had 7 balls of which 3 were red, 2 were green and the others different. The number of arrangements came out to be \( \frac{7!}{3!2!} \)

Now, let’s generalize. Suppose there are n objects, of which ‘p’ objects are identical and of a particular type (e.g. colors in case of balls), and ‘q’ objects are identical and of another type, and the remaining are distinct, then number of arrangements of these n objects in a row, will be \( \frac{n!}{p!q!} \)

Further, if ‘r’ objects are also identical and of a third type, then the number of arrangements will change to \( \frac{n!}{p!q!r!} \), and so on.

Lesson Summary

  1. The number of ways of arranging n objects in a row, of which p objects are identical, and of type 1, q other objects are identical and of type 2, ‘r’ other objects are also identical and of type 3, and the remaining are distinct, is given by \( \frac{n!}{p!q!r!} \)

That’s it for this lesson. I’ll cover a few examples related to this concept in the next lesson.

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