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Using the $26$ English letters, the number of $5$-letter words that can be made if the letters are distinct is determined as follows:

$26P5=26\times25\times24\times23\times22=7893600$ different words.

What if the letters in each word are in alphabetical order?

For example, the word JLOQY is valid, but the word JUMPY is invalid since U can not be before M

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  • $\begingroup$ Do you allow repeats for the main question (e.g. ABBEY)? $\endgroup$
    – Parcly Taxel
    Jul 6, 2019 at 11:42
  • $\begingroup$ @ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5. $\endgroup$
    – Hussain-Alqatari
    Jul 6, 2019 at 11:44
  • $\begingroup$ As an aside, I prefer to think of the result as a binomial coefficient $\binom{26}{5}$ rather than as a falling factorial divided by a factorial $\frac{26\frac{5}{~}}{5!}$, or using your notation $\frac{~_{26}P_5}{5!}$ $\endgroup$
    – JMoravitz
    Jul 6, 2019 at 11:56

3 Answers 3

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Hint. How many ways can you choose the five different letters? Once you have them, in how many ways can you organize them in alphabetical order?

(This assumes the letters are distinct.)

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    $\begingroup$ So will it be $\frac{26P5}{5!}=\frac{7893600}{120}=65780$ different words? $\endgroup$
    – Hussain-Alqatari
    Jul 6, 2019 at 11:46
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Ethan Bolker
    Jul 6, 2019 at 11:54
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Divide out the number of permutations of five letters ($5!$), since only one is correct.

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The different number of ways to select 5 alphabets from 26 alphabets= $26C5$.

Arrange the alphabets each collection in the required order.

Thus the answer is $26C5$.

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