Combinations in alphabet

BrusselSproot

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Hello.

How many 3 letter combinations would there be in the alphabet? They’re can’t be repeating combinations (ADG, GDA, DAG) it doesn’t matter what order they are in, there just can’t be the same 3 letters in any orientation in a combination of 3 letters.

Hope this makes sense, thank you!
 
How many 3 letter combinations would there be in the alphabet? They’re can’t be repeating combinations (ADG, GDA, DAG) it doesn’t matter what order they are in, there just can’t be the same 3 letters in any orientation in a combination of 3 letters.
These are a simple combination of twenty-six items taken three at a time: \(\displaystyle \dbinom{26}{3}\) SEE HERE
 
Hello.

How many 3 letter combinations would there be in the alphabet? They’re can’t be repeating combinations (ADG, GDA, DAG) it doesn’t matter what order they are in, there just can’t be the same 3 letters in any orientation in a combination of 3 letters.

Hope this makes sense, thank you!
Your wording is awkward, so let's clarify it.

When you say, "there can’t be repeating combinations (ADG, GDA, DAG) it doesn’t matter what order they are in", you mean that you want to count distinct combinations, ignoring order. "Repeating" sounds as if you meant that no letter within a combination can repeat, but that appears to be covered by your next clause instead, "there just can’t be the same 3 letters in any orientation in a combination of 3 letters". But this, taken literally, would only exclude, say, AAA, in which all letters are the same, and not cases like AAB where only two letters are the same.

But it does appear that pka is answering your intended question; "combinations", as a technical term, means distinct subsets, that is, sets of (3) letters with no repetitions, ignoring order.

For information about the concept and how it is calculated (and, more important, why), see https://www.mathsisfun.com/combinatorics/combinations-permutations.html; for a more formal explanation, see https://en.wikipedia.org/wiki/Combination.
 
There are 26 choices for the first letter, then, since letters cannot repeat, 25 choices for the second letter, and 24 choices for the third letter. That would give \(\displaystyle 26(25)(24= \frac{26!}{23!}\) non-repeating combinations. Since you don't want to count "ADG", "GDA", "DAG" (and, I presume, "AGD", "GAD", and "DGA") as separate, divide by 3!. That gives \(\displaystyle \frac{26!}{23! 3!}\). That is the "binomial coefficient", also denoted \(\displaystyle \begin{pmatrix}26 \\ 3\end{pmatrix}\) or \(\displaystyle \begin{pmatrix}26 \\ 23\end{pmatrix}\).
 
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