Arrangements. Arrangements with repetition

Arrangements $A_n^k$ without repetition

Choose $k$ distinct elements from $n$ and place them on positions (1st, 2nd, …, $k$ -th) — $k\leq n$ . $A_n^k=n(n-1)(n-2)\cdots(n-k+1)=\dfrac{n!}{(n-k)!}.$ With repetition, under independent steps with “return to stock”, each of $k$ positions has $n$ options, giving $\tilde A=n^k$ (a “code of length $k$ from $n$ symbols”).

Remember the product of $k$ decreasing factors $n(n-1)\cdots$ for the no-repetition case

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