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Assuming the probability of getting an object opening a pack of a collection is uniform, what is the probability of getting 14 objects without a single duplicate if the collection is made of 24 objects?

I would like to know if my execution is correct:

$n = 24$
$k = 14$
$\binom{a}{b}$ is binomial coefficient

Total cases are the combination of 14 objects selected from 24 with repetition

Total cases = $\binom{n+k-1}{k}$ = 6.107.086.800

Good cases are the combination of 14 objects selected from 24 without repetition

Good cases = $\binom{n}{k}$= 1.961.256

$p(\text{E = no duplicates in 14 packages}) ={\text{good cases}\over\text{total cases}}=3.21144\times10^{-4}$ which is then 0.0321144%

is it correct?

Lucenaposition
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Gabriele Passoni
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1 Answers1

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The combination with repetition formula does not generate equiprobable outcomes.

You can instead imagine that a $24$ faced die is thrown $14$ times, and you want the probability that each outcome is unique.

So you could just multiply out $\frac{24}{24}\frac{23}{24}\frac{22}{24}...\frac{11}{24}$,

more conveniently written as $\Large\frac{{}^{24}P_{14}}{24^{14}}$

true blue anil
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  • Thanks for the detailed explanation. The doubt came up to my mind just exactly thinking as your example (24 faced die). How would it be to solve it using my combinatory (so good cases/total cases) approach? – Gabriele Passoni Sep 28 '23 at 10:46
  • Well, by the combinatorial approach, I suppose you mean $\Large\frac{\binom{24}{14}14!}{24^{14}}$, which is exactly the same as $\Large\frac{{}^{24}P_{14}}{24^{14}}$ – true blue anil Sep 28 '23 at 12:27
  • Am I wrong or basically the solution is made using dispositions instead of combinations? – Gabriele Passoni Sep 28 '23 at 17:13
  • Yes, permutations where the number picked is less than the total number of objects. – true blue anil Sep 28 '23 at 17:34