[Math] The 5 Reindeer – Permutations & combinations

combinationspermutations

You need to put your reindeer, Gloopin, Balthazar, Bloopin, Prancer, and Quentin, in a single-file line to pull your sleigh. However, Prancer and Balthazar are best friends, so you have to put them next to each other, or they won't fly.

How many ways can you arrange your reindeer?

I know there is another question asking the same thing, but my question is not what that person was asking. Mine is different.

I know and understand this:

  1. So, the total number of unique choices we could make to get to an
    arrangement of reindeer is $5⋅4⋅3⋅2⋅1 = 120$. Another way of writing
    this is 5! or 5 factorial. But we haven't thought about the two
    reindeer who have to be together yet.

  2. We can count the number of arrangements where Prancer and Balthazar
    are together by treating them as one double-reindeer. Now we can use
    the same idea as before to come up with $4⋅3⋅2⋅1 = 24$ different
    arrangements. But that's not quite right.

  3. Why? Because you can arrange the double-reindeer with Prancer in
    front or with Balthazar in front, and those are different
    arrangements! So the actual number of arrangements with Prancer and
    Balthazar together is $24⋅2 = 48$

I do not understand why you have to multiply 24 by 2 to get the answer. Can someone help me understand this?

Best Answer

It is because when you consider the reindeer next to each other you get $4!$, but this has not been generalized, because the order Prancer and Balthazar can be switched to Bathazar and Prancer, and still satisfy the requirement of them being next to eachother.

For instance, let Prancer, Balthazar, Bloopin, Gloopin, and Quentin be one combination of the $4!$ combinations. But Balthazar, Prancer, Bloopin, Gloopin, and Quentin are also a valid combination but you only counted one of these two combinations.

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