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[Combinatorics] Ordering vs creating strings coincidence

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They're related by a stars-and-bars argument. Think of the a's as stars and b's as bars, then abaaba → abbc.
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> How many 4 length strings can we make out of the letters a, b, and c?

That's 3^4 because "strings" implies order matters. If you only care about *selections*, then yes it's 6C2 because of stars and bars. You're choosing two places (out of 6) for the bars, while with aabbbb you're choosing two places for the a's out of 6.

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