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1. How many permutations of the 26 letters of the alphabet are there that contain none of the sequences MATHS, JANE, NIGEL, or DUR??

the are 26! total permutations, and (26-r+1)! arrangements of the letters when they contain a sequence of length r.

So S1 = 24!+23!+2.22! as this takes into account all of the sequences. (With MATHS having r = 5, JANE r=4 etc).

Now I get this much. However the solution goes on to say that you must take into account the fact that the strings may overlap so you get:

S2 = 21! + 2.22! + 18!,
S3 = 16!
S4 = 0

So the overall answer is: 26! - S1 + S2 - S3 +S4.

However despite understanding that inclusion-exclusion must be used, I need to know how they got S2, S3 and S, and would like to know what they represent. I don't understand where the sums came from for those...
2. S2 = # of permutations containing MATHS & JANE or MATHS & NIGEL or MATHS & DUR or ...(You should have 6 pairs of words, but note some of them are not possible).
S3 = # of permutations containnig MATHS + NIGEL + DUR (only way you can have 3 words)
3. (Original post by DFranklin)
S2 = # of permutations containing MATHS & JANE or MATHS & NIGEL or MATHS & DUR or ...(You should have 6 pairs of words, but note some of them are not possible).
S3 = # of permutations containnig MATHS + NIGEL + DUR (only way you can have 3 words)
Whats wrong with having MATHS + JANE + NIGEL??
4. You can't have Maths + Jane; you'd need two 'A's.
5. Ahh I see, thanks...

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Updated: July 28, 2010
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