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## Homework Statement

Given the word "DEEPBLUE" - possibilities of words such that two "E"-s are not adjacent in any word.

## Homework Equations

## The Attempt at a Solution

There are 8 symbols. If 2 E-s cannot be adjacent then how can I calculate the possibilities of placing all the 3 E-s such that the criterion is satisfied?

A small side-tracking - in a way it reminds me a little bit of a problem where we have a long-table with N seats and we have N/2 men and women, what are all the possibilities of seating people such that no 2 men or 2 women sit next to one another? Well it would be 2(N/2)! - rule of product.

The E-s are all identical now, therefore there is no need for permutations.

ALL of the possible combinations of 3 would be C

_{(8,3)}, but this also considers the possibilities where all 3 are together AND when 2 are adjacent and the 3rd is not adjacent to either side.

In case of 3 E-s stuck together there are only 6 possibilities, right? We can subtract that.

What about the number of possibilities of 2 E-s adjacent and the 3rd not adjacent?

If 2 E-s are together then there are 7 possibilities to place the 2 E-s and then:

2 of possibilities have the 2 E-s at the side, wihch leaves 5 possibilities to the leftover E to not be adjacent - 10 possibilities

5 possibilities where for each there are 4 possible positions for the left over E - 20 possibilities * 2 (counting also the mirror image) - 40 possibilities.

Total 50 possibilities that violate the criterion hence [itex]\frac{8!}{5!3!} - 50[/itex] possibilities. 6 possibilities to place the E-s.

We have 5 unique symbols left over and 5 spots - 5! possibilities.

Applying the rule of product the total number of words would be 720. Is this correct?

Expanding on the second part of the problem.

What happens if for 5 left over spots, we have 3 unique symbols and for now, no criteria be set on any repetition? is it simply 3

^{5}possibilities?

If there was extra conditions, such that if the left over symbols are A,B,C and ANY of the symbols are not allowed to be adjacent.

I would have 3 possible ways to place the ABC in the 5 symbol space. Then also 6 ways to rearrange ABC - 18 possibilities.

This is where my wits end:

For any arbitrary variation within the 5 symbol space, how do I exactly know how I can place the other symbols such that the criterion holds.

In the context of the original problem - the 5 symbol space is not all in one piece, how do I account for the disjuctions created by the E-s that will possibly increase the number of possible words?

What kind of material is there to look into to get a better grasp for this type of problem?

*Is there a way to generalize this to N symbols and M repeative symbols?*