B Probability of arranging identical objects

[MeesaWorldWide]

TL;DR Summary

If all the letters of the word NUNAVUT are randomly rearranged, what is the probability that there will be an N at each end?

Total ways to arrange the 7 letters: 7! / (2! x 2!) = 1260

Ways to have an N at each end: N _ _ _ _ _ N
There are 5 other letters in the middle, and two of them repeat (U), so the middle 5 are found by 5! / 2! = 60

Now, here is where I am unsure what to do. Since the N's are identical, do we distinguish between them? If we switched both the N's, there is still an N at either end, but would that be considered a separate case? If so, we do 60 x 2 = 120.
Then the probability of an N being at each end is 120 / 1260 = 0.095

If not, then the probability is simply 60 / 1260 = 0.048

Any insight here would be appreciated. 0.095 seems a bit high to me, but not treating N _ _ _ _ _ N and N _ _ _ _ _ N as different cases also makes sense...

 

[anuttarasammyak]

The word has letters
N N
U U
A
V
T
It is not clear two N and two U are distinguished or not in our count. If we distinguish two N and two U, for an exaple by capital and lower case
N n U u A V T

I think the rule to count the cases should be shared for the investigation.

As for N n U u A V T rule, number of all the cases : 7!
N is at one edge and n is at another edge : 5! x 2

So the probality is 1/21

 

[PeroK]

There's no need for counting in this case. The probability that the first letter is N is $\frac 2 7$. And, the probability that the last letter is N, given the first letter is N, is $\frac 1 6$. The probability of both is the product of these, hence $\frac 1 {21}$.

 

[PeroK]

PS note that calculation applies to any two chosen positions for the N's. The probability that the first two letters are both N is the same, by the same calculation. Likewise the first and third letters; or the third and sixth etc.