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$^{n}P_r$, the number of permutations of $n$ distinct objects taken $r$ at a time, equals:
A$\dfrac{n!}{r!(n-r)!}$
B$\dfrac{n!}{r!}$
C$n \cdot r$
D$\dfrac{n!}{(n-r)!}$
Answer & Solution
Correct answer: D. $\dfrac{n!}{(n-r)!}$
$^{n}P_r = \dfrac{n!}{(n-r)!} = n(n-1)(n-2)\ldots(n-r+1)$, which is $r$ consecutive descending factors starting from $n$.
Derivation: choose 1 of $n$ for the first slot, then 1 of $(n-1)$ for the second, and so on through $r$ slots. The product is the descending factorial.
Option C is $^{n}C_r$ (combinations, where order doesn't matter). The difference between $P$ and $C$ is the $r!$ in the denominator, which divides out arrangement-order redundancy.
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