Combinatorics: Distinguishable and Indistinguishable Variables

Question

$3$ men and $5$ women (each of the $8$ being different from all the rest) are lined up for a photograph. Also in the line are $3$ identical armadillos which are completely indistinguishable from each other.

a) How many ways can the $8$ humans and $3$ armadillos be lined up for the photo?

b) How many ways can they all line up if none of the men are adjacent to each other?

I think I know how to do this if there weren't the armadillos involved, I know the fact that the armadillos are indistinguishable is important. From my understanding I think it indicates that in includes a combination calculation as the items are indistinguishable? But otherwise, I do not know how to calculate this!

Answer 1

Let's first deal with a)

I tend to be an equal opportunist for all species, so let's treat all humans and armadillos as different to start.

In that case, it's as if we have to arrange 11 humans in a line, which would lead to 11! possibilities.

Now of course, not everyone treats armadillos so nicely. We must account for the fact that armadillos are indistinguishable. How? Let's think about how many times we overcounted.

In our first estimate of the number of combinations, assume that armadillo 1, 2, 3 are placed at positions x, y, and z respectively. Note that if I place armadillo 1 at y, 2 at z, and 3 at x, the combination should be treated identically. So we have overcounted a number of times equal to the number of ways to arrange 3 armadillos among each other, which is 3!.

So, the correct answer to part A is $\frac{11!}{3!}$.

In part B, the armadillos can be treated in an identical fashion.

Answer 2

Imagine the armadillos were distinguishable. You mentioned you are then able to solve the problem; solve it, and then divide by $6$.

Why? Because the arrangements you are now counting come in groups of $6$, differing only in the arrangement of the armadillos. Taking $\frac{1}{6}$ of the result would tell you how many genuinely different arrangements exist when the armadillos are deemed identical.

Answer 3

The other answers do a fine job of describing the thought process when you wish to use the technique of "dividing by symmetry."

I find it helpful to also take the time to describe how to accomplish the same thing without needing to resort to that.

Let us approach via multiplication principle:

• First, pick which three of the $11$ locations are occupied by the armadillos and place them there. This can be accomplished in $\binom{11}{3}$ ways.
• Next, in the remaining $8$ locations, arrange the humans there. This can be accomplished in $8!$ ways.

There are then a total of $\binom{11}{3}\cdot 8!$ different ways to arrange the people and animals for a picture.

You should notice that $\binom{11}{3}\cdot 8! = \frac{11!}{3!8!}\cdot 8! = \frac{11!}{3!}$, and is the same result as shown in the other answers, just written in a different way.