What is the permutation of choosing the just 3 balls in a pool of 16 balls?

Question

I was reading permutation without repetition which says that
In a pool if $16$ balls there are $N!$ possibilities, which is,
$16 × 15 × 14 × 13 × ... = 20922789888000$ possibilities
But when we don't want to choose them all, just $3$ of them, and that is then:
$16 × 15 × 14 = 3360$ or $frac {16!}{13!}$
My doubt is when we want to choose only $3$ balls out of $16$ then it is $3! = 6$. Is this understanding correct?

Commander Vimes · Answer

I assume that order matters. Then the correct answer is $16 times 15 times 14$, because for the first ball I have 16 possible choices, then for the second ball, when one is removed, I have 15 choices (i.e for each first ball chosen I have 15 options remaining). And then, for the third ball, I have 14 options remaining.
$3!=6$ would be correct if I had total of 3 balls. For example, I have balls a, b, c. Then possible choices are
abc
acb
bac
bca
cab
cba
But if I add fourth ball here, d, you will see that I have to consider other possibilities also, for example, acd.

What is the permutation of choosing the just 3 balls in a pool of 16 balls?

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