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Barney Stinson, attempting a perfect week, wants to spend a whole week going on a date with different girls. He will go out on the nights of each day in the week (Monday through Sunday) choosing from the girls: Nora, Quinn, Abby, Wendy, Shannon, Shelly, Honey, Kelly, Jenkins, Stacey, Meg and Karina.

Barney does not want to take the same girl to bed more than once in the week. How many possible outcomes are there?

At first I thought that the answer is $\binom{12}{7} = \frac{12!}{(12-7)!7!}$. However, I was told that the order matters. So it is $7!\binom{12}{7} = \frac{12!}{(12-7)!}$?

The problem says "choosing from" therefore order must not matter, or does it?

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"Choosing from" seems to be used here with the meaning of "picking" or "selecting", rather than the mathematical choice function. The order is implied to matter, though it's not quite clear; I prefer to ask for clarification on this sort of thing when possible.

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  • $\begingroup$ Do you mean that whether or not order matters is something that we cannot deduce? It must be stated by the professor or perhaps Barney himself? $\endgroup$ – user198044 Mar 4 '16 at 15:26
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    $\begingroup$ This specific problem isn't worded clearly enough to deduce whether order matters or not. $\endgroup$ – DylanSp Mar 4 '16 at 15:36

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