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In the Khan Academy resource on "Combinations," the following question appears:

Five reindeer:

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?

In the hint portion, we're given the following explanation:

  • 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.
  • 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

The bolded part above is what's confusing to me.

If we assume five spaces for the reindeer: ----- then it would be 5*4*3*2*1 or (5!) for 120 total arrangements.

And if we're putting P and B together, we would have either PB--- or BP---.

What I'm confused about is how this is 1*4*3*2 and not 1*1*3*2*1. Or, wouldn't it be (2!)*(3!).

Thanks!

Steve Byer
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    Why do you think it would be $11321$? Or $2! * 3!$? The intuition behind that thought is still not clear. By what you have written, don't forget that possibilities like $PB _$ and $_BP _ _$ exist. – Sarvesh Ravichandran Iyer Aug 20 '16 at 06:00
  • An alternate way, if you don't like the "double reindeer" bit... have gloopin, bloopin and quentin order themselves. Not in the slots... we don't care about slots at the moment, just have the three of them arrange themselves in a line. This can be done in $3!$ ways. Now, let Prancer decide where he wants to insert himself into the other reindeer. This can be just before bloopin, just before gloopin, just before quentin, or after the last one. This can be done in $4$ different ways. Now, let Balthazar decide if he is before or after Prancer: 2 ways. for a total of $3!\cdot 4\cdot 2$ ways. – JMoravitz Aug 20 '16 at 07:03

3 Answers3

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First of all, combinations with restrictions are called permutations :)

A neat way to do such problems is to draw as many boxes as there are necessary 'team members' and pencil in the number of permutations for each member. In doing so, you could also draw an arrow protruding from each box where you list the constituent permutations.

In this case, a few sets of boxes can be used as the problem can be split into disjoint cases.

(Since I don't know how to draw arrows on this site, I'll list the constituent permutations inside the boxes :)

The first case: $\fbox{2 (P. or Ba.)}\fbox{1 (Ba.)}\fbox{3 (G., Bl. or Q.)}\fbox{2 (Bl. or Q.)}\fbox{1 (Q.)}$

Of course, the reindeers could be in a different order within the restrictions of the question — but the same amount of permutations would result. For purposes of visualisation, you can arbitrarily assume which member would go where (within the restrictions of the question).

Make sure, of course, to put the restrictions in first as the permutations for each other position in the 'team' will depend on the restrictions — in this case that Prancer and Balthazar must be next to each other.

The amount of permutations for Case 1? $2*1*3*2*1 = 12.$

There are three other disjoint cases — where Prancer or Balthazar are next to each other in 2nd and 3rd position, 3rd and 4th, or 4th and 5th.

The total number of permutations (the sum of the permutations for each disjoint case) $=48$.

Try it out using this method and see how you go :)

Mad Banners
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  • Treating Prancer and Balthazar as one double-reindeer, I believe you just replace their two positions with one :) @Steve Byer – Mad Banners Aug 22 '16 at 23:47
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We can see that we have 5 reindeers, among them Gloopin,Bloopin and Quentin can come in any sequence but 'Prancer and Balthazar' has to come together. So we can group them like below Gloopin,Bloopin,Quentin,(Prancer,Balthazar). Considering this condition we can put (Prancer,Balthazar) in spaces(__) like. __Gloopin__Bloopin__Quentin__.Now we have 4 places where we can put both. So number of possible ways we can arrange them is 4!

Another thing we need to consider is we can interchange places of (Prancer,Balthazar) like (Balthazar,Prancer) so we have 2! possible ways to do this as there are 2 members.

Finally, we can conclude that we can arrange reindeer with 4!*2! i.e(4*3*2*1) * (2*1) = 48

Prasanna
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I can't understand why you are thinking of the alternatives you have in mind, so I will just try to clarify the given (correct) approach.

The two patterns to permute are $\boxed{BP}\; G\;B\;Q$ and $\boxed{PB}\; G\;B\;Q,$ hence ans $=2\cdot4!$

true blue anil
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