# How many 5-letter words can we make if the letters are in order?

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

• Do you allow repeats for the main question (e.g. ABBEY)? – Parcly Taxel Jul 6 '19 at 11:42
• @ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5. – Hussain-Alqatari Jul 6 '19 at 11:44
• 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 $\frac{~_{26}P_5}{5!}$ – JMoravitz Jul 6 '19 at 11:56

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.)

• So will it be $\frac{26P5}{5!}=\frac{7893600}{120}=65780$ different words? – Hussain-Alqatari Jul 6 '19 at 11:46
• Yes, that's right. – Ethan Bolker Jul 6 '19 at 11:54

Divide out the number of permutations of five letters ($$5!$$), since only one is correct.

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 $$26C5$$.