I am trying to understand the answer to the following problem:
The eleven letters of the word E-X-A-M-I-N-A-T-I-O-N are written on eleven separate
pieces of card. Find the probability that the four letter word E-X-A-M will appear in one of
these eleven letter arrangements.
My attempt was as follows: EXAM may be considered as 1 "unit" so there are 8 spaces it can move in to around the other 7 letters. There are 2 I s and 2 N s, and 7 letters which can be moved so the number of possible arrangements is $\frac{7!}{2!2!}$ x 8 = 10,080 however the solution gives 5040 saying that swapping the 2 A s makes no more arrangements, so there are 3 pairs of duplicates, not 2. (I am not worried about getting the probability at this point because it is very, very easy to do once you have the number of arrangements)
I couldn't really understand this so I made up the following smaller example:
Find the number of arrangements of the word "THREE" with the letters "HRE" together.
Treat "HRE" as 1 "unit" again and move it around the 3 available spaces .The remaining letters T and E can be swapped to produce another solution, for example: HREET and HRETE. Hence we have 3×2! = 6 solutions. According to the logic of the solution for the examination question, we should have $\frac{2!}{2!}$ x 3 solutions because of the double E but that is untrue:
1.HREET
2.HRETE
3.THREE
4.EHRET
5.ETHRE
6.TEHRE
What am I missing?
Best Answer
There must be some error in the solution. Your logic is absolutely correct. When “EXAM” is a unit, then “EXAM” is not similar to “A”. Hence, swapping two "A"s is no longer possible.
Your "Find the number of arrangements of the word "THREE" with the letters "HRE" together." is an example of it.