Calculating the Probability of Combinations of Reindeer Arrangements

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The discussion focuses on calculating the probability of valid arrangements of reindeer, specifically addressing the constraint that reindeer with "r" in their names cannot be adjacent. With 8 reindeer, 4 containing an "r", two configurations are identified: x R x R x R x R and R x R x R x R x. The total valid arrangements are calculated as 1152, which is adjusted to 1138 by excluding arrangements where Blitzen and Donner are adjacent. The final probability of valid arrangements is determined to be P(X) = 569/20160.

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View attachment 6277 Is there an easy way to do this?
 

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I think what I would do is begin with the rule that no reindeer with"r" in their names may be adjacent. We see that of the 8 reindeer, 4 of the have an "r" in their name. If we denote a reindeer with an "r" in their name with an R and all others with an x, we see we have two possible configurations:

x R x R x R x R

R x R x R x R x

How many ways can we arrange the reindeer in either of these two ways?
 
4! for each one with r, times 4! for each combination of those without r, times 2 for the opposite patterns

4!*4!*2=1152?
 
Yes, that's correct. Next, let's look at the restriction that Blitzen and Donner cannot be adjacent. We need to find out how many ways the can be adjacent and subtract that from 1152. Let's consider one of the arrangements:

x R x R x R x R

There are 4 places Blitzen could be (where the x's are). How many places can Donner be such that he is next to Blitzen?
 
is it 7?
 
Okay, then how many arrangements of this type:

R x R x R x R x

do we also exclude?
 
7 as well?
 
Ilikebugs said:
7 as well?

Correct! (Yes)

So that means we have:

$$1152-(7+7)=1138$$

ways to legitimately arrange the reindeer. Next, we need to find out the total number of ways the reindeer can be randomly arranged...(Thinking)
 
8! or 40320?
 
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So, you now have enough information to answer the question, as the probability requested is the ratio of the good arrangements to the total arrangements:

$$P(X)=\frac{1138}{40320}=\frac{569}{20160}$$
 

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