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Confused about permutations... watch

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    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...
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    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)
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    (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??
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    You can't have Maths + Jane; you'd need two 'A's.
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    Ahh I see, thanks... :top:
 
 
 
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