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A word is defined as a nonempty (possibly meaningless) sequence of letters. How many $6$-word sentences can be made using each of the $26$ letters of the alphabet exactly once? Generalise the result to the number of sentences consisting of $w$ nonempty words using exactly once each letter from a $l$-letter alphabet.

My Attempt We have $^{26}P_{26}$ ways of arranging the $26$ letters. Then, in order to have six words there have to be five spaces. For each of the previous arrangements there are $^{25}C_{5}$ ways to separate groups of letters into six. Thus, the answer is $^{26}P_{26}\,\cdot\,^{25}C_{5}\approx2\times10^{31}$.

For the generalisation we have $^{l}P_{l}\,\cdot\,^{l-1}C_{w-1}$.

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    $\begingroup$ This is correct. It is a nice application of the stars and bars technique. $\endgroup$
    – TravisJ
    May 18 '15 at 17:56
  • $\begingroup$ @TravisJ Ah, that didn't occur to me. But only for the $^{25}C_{5}$ part? $\endgroup$ May 18 '15 at 18:03
  • $\begingroup$ Yes, the method of breaking up the 26 letter word into smaller words. I should point out that this is correct provided that you allow words to have no letters. $\endgroup$
    – TravisJ
    May 18 '15 at 18:29
  • $\begingroup$ The second part should be 5 out of 25, otherwise you will be getting combinations with less than 6 words $\endgroup$
    – DannyDan
    May 18 '15 at 19:32
  • $\begingroup$ @DannyDan What are you referring to exactly please? Thanks. $\endgroup$ May 19 '15 at 4:59

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