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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?

Arjun
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mahesh87
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  • Not sure what you are asking. You give two different values for the number of ways to choose $3$ (numbered, I'm guessing) balls. The first one is correct IF you take order into account (so that drawing $(1,2,3)$ in that order is different than choosing $(1,3,2)$). The second is not correct at all. You should easily be able to list more than $6$ ways to make the selection. – lulu Jul 23 '20 at 11:37
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    after few more analysis I understood that (16!/(16-3)!) which says that 3 balls can be any of those 16 pool balls, so the out of all the permutations the first 13! cancelled out which leaves to 3,360. was going thru this link https://www.mathsisfun.com/combinatorics/combinations-permutations.html – mahesh87 Jul 23 '20 at 11:39
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    The remaining 13 balls will still be arranged -- their permutations just don't matter to you, so you divide them out of the product. – Andrew Chin Jul 23 '20 at 11:50

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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.

Commander Vimes
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