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Consider the first ten letters of the alphabet $\{A,B,C...J\}$ and consider any three letter sequence a word. How many three letter 'words' can be constructed from this set in which all the letters are different and in which the letters are in alphabetical order.

It is easy to see that there are 720 words that can be made from three different letters. I have been informed that imposing the second constraint reduces this to 120 words, but it is not clear to me why exactly $\frac{1}{6}$ of the of the first set are in alphabetical order. Any argument that clarifies this relationship would be greatly appreciated.

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The answer is ${10\choose 3} = 120$, because you must choose distinct letters, and for any set of three distinct letters you only get to construct one word based on alphabetical order.

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  • $\begingroup$ The answer that you provided isn't initially intuitive but is quite simple and helpful. $\endgroup$ – Matthew Anderson Apr 10 '17 at 4:03
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If you have any three different letters, and you represent the 'lowest' letter (i.e. the one that comes first in the alphabet) as L, the 'highest' as H, and the one on the middle as M, then you have 6 possibilities, only the first one of which is in alphabetical order:

LMH

LHM

MLH

MHL

HLM

HML

So: 1 out of 6

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