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Suppose I have a list of $2000$ words and I want to combine them, randomly choosing a word from this list each time

e.g. Apple-Orange-Boat or Apple-Apple-Apple

How many possible combinations are there?

Is it simply $2000^3$?

ycomp
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  • It depends whether the order matters. For example, are you counting Apple-Orange-Boat and Orange-Apple-Boat as different things? – Sam Weatherhog Oct 15 '15 at 00:28
  • yes, different.. I'm using them as "fairly" unique human-readable identifiers, just like Stanley Carol and Carol Stanley would be different people, so too are these different. – ycomp Oct 15 '15 at 00:30
  • In that case, since you are allowing repeats and the order matters the total number of combinations is $2000^3$. You have 2000 choices for the first word, 2000 for the second, and 2000 for the third. – Sam Weatherhog Oct 15 '15 at 00:34

2 Answers2

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No. If you want a combination, then "Apple-Orange-Boat" is the same as "Orange-Boat-Apple". The order does not matter and therefore $2000^3$ is counting too much. For combination with repetitions, your answer is $$\begin{pmatrix}n+k-1\\k-1\end{pmatrix}=\begin{pmatrix}2000+3-1\\3-1\end{pmatrix}=\begin{pmatrix}2002\\2\end{pmatrix}.$$

Of course, if you want to consider the order, then you are not talking about combinations. In this case, the answer would be $2000^3$.

João Victor Bateli Romão
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  • yes I did not mean the mathematical term "combinations", but the colloquial term - it has been many years since I studied any math – ycomp Oct 15 '15 at 01:20
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Without ordering and without repetitions, eg Apple-Orange-Boat is the same as Orange-Apple-Boat and Orange-Orange-Orange is not allowed:

$$ {n \choose k} = {n \choose n-k} = \frac{n!}{k!(n-k)!} = \frac{2000!}{3!(2000-3)!} = 1.331334E9 $$

Without ordering and with repetitions, eg Apple-Orange-Boat is the same as Orange-Apple-Boat and Orange-Orange-Orange is allowed:

$$ {n+k-1 \choose k} = {n+k-1 \choose n-1} = \frac{(n+k-1)!}{k!(n-1)!} = \frac{(2000+3-1)!}{3!(2000-1)!} = 1.335334E9 $$

With ordering and without repetitions, eg Apple-Orange-Boat is different from Orange-Apple-Boat and Orange-Orange-Orange is not allowed:

$$ \frac{n!}{(n-k)!} = \frac{2000!}{(2000-3)!} = 7.988004E9 $$

With ordering and with repetitions, eg Apple-Orange-Boat is different from Orange-Apple-Boat and Orange-Orange-Orange is allowed:

$$ n^k = 2000^3 = 8E9 $$

So, based on your comments, your instinct was indeed correct.

cr3
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