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As an example, if I have 10 items and am choosing 2, the equation $\frac{n!}{(n-k)!k!}$ will result in 45.

However, intuitively there are 10 items to choose from and then 9 leftover to choose from again, so the result would be 10*9=90. (Equation would be $\frac{n!}{(n-k)!}$)

How do I know which of these to use in what scenario?

RobPratt
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ikswa
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    Sure, if the order of the two elements matters, the correct answer is $90.$ But if it doesn’t? – Thomas Andrews Feb 11 '22 at 17:51
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    The $k!$ comes from disregarding the "order' of the elements you choose. You can think of it in this way: the number of ordered $k$-tuples with distinct elements I can form with the set ${a_1, \dots, a_n}$ ( $k \leq n$) is $\frac{n!}{(n-k)!}$, while the number of sets with $k$ elements I can form is $\frac{n!}{(n-k)!k!}$. – user480840 Feb 11 '22 at 17:52
  • Dividing by k! removes the number of permutations of k objects chosen, when order of the objects does not matter. – amWhy Feb 11 '22 at 17:56
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    Does this answer your question? – J.G. Feb 11 '22 at 17:56
  • Nice find, @J.G. ! – amWhy Feb 11 '22 at 17:58

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