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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 at 11:42
  • $\begingroup$ @ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5. $\endgroup$ – Hussain-Alqatari Jul 6 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 at 11:56
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Hint. How many ways can you choose the five different letters? Once you have them, now 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 at 11:46
  • $\begingroup$ Yes, that's right. $\endgroup$ – Ethan Bolker Jul 6 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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