- #1

- 1,331

- 40

If I have a set of 13 elements, I can arrange that 13! different ways, because Psub(13,13) = 13!/0!.

If I pick 2 elements from those 13 elements, I can get 13!/11! different results.

Is that what it means?

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- Thread starter 1MileCrash
- Start date

- #1

- 1,331

- 40

If I have a set of 13 elements, I can arrange that 13! different ways, because Psub(13,13) = 13!/0!.

If I pick 2 elements from those 13 elements, I can get 13!/11! different results.

Is that what it means?

- #2

- 36,029

- 6,705

Imagine the 13 elements in a row, and suppose the leftmost two will be the two picked. There are 13! orderings altogether. For a given pick of two, there are 2!*11! orderings that lead to it. So the number of such pairs is 13!/(2!*11!).

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