#### Swazination

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- Thread starter Swazination
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Now suppose we had two letters and two slots. How many combinations would there be then? Thinking about it, we can force the first slot to be A, then there's two possibilities for the second slot. Similarly, if we force the first slot to be G, we have two additional combinations. So we have two possibilities for the first slot, and for each of those two possibilities, we have two possibilities for the second slot. That gives us a total of 2 * 2 = 4 = 2

Now suppose we had two letters and three slots. How many combinations would there be then? I see that we can arbitrarily fix the first two slots and then let the third one vary. Since we already know that there's four possible ways to arrange the first two slots, and for each of those four possibilities, there's two choices for the third slot, that gives us a total of 4 * 2 = 8 = 2

Now let's suppose we had three letters (A, G, and C) and two slots. By the exact same logic as before, we have three choices for the first slot, and for each of those three choices, we have three choices for the second slot. That gives us a total of 3 * 3 = 9 = 3

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In the post you use the term

NOTE that

If we want the number of ways to put \(\displaystyle \bf{N}\) identical items (in this case five choices) into \(\displaystyle \bf{k}\) distinct cells (in this case four letters) is \(\displaystyle \dbinom{N+k-1}{N}\)

(in this case \(\displaystyle \dbinom{5+4-1}{5}=\dfrac{8!}{5!\cdot 3!})\)

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