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Legen Sie den Lieferort fest, um Produkte entsprechend Ihrer Wahl anzuzeigen.
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Ausgewählter Laden
Skroutz Käuferschutz
Die Preise werden berechnet für:Deutschland, Andere Zahlungsoptionen
Figur Scarlet Witch mit Wackelkopf aus der Comicserie The X-Men von Marvel.
Spezifikationen werden von offiziellen Hersteller-Websites gesammelt. Bitte überprüfen Sie die Spezifikationen, bevor Sie Ihren endgültigen Kauf tätigen. Wenn Sie ein Problem bemerken, können Sie melden Sie es hier
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genau wie auf dem Foto
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The besttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt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I have a few questions about the 2nd part of the question, the one that asks for the number of ways to distribute $n$ indistinguishable objects into $k$ distinguishable boxes, where each box can contain at most $m$ objects.
I think the answer is $\binom{n+k-1}{k}$, but I'm not sure how to prove it. I know that the number of ways to distribute $n$ distinguishable objects into $k$ distinguishable boxes is $k^n$, but I'm not sure how to use that here.
Biko03 2019-02-25: Let us consider the number of ways to distribute the $n$ objects into the $k$ boxes such that no box is empty.
To do this, we can use the principle of inclusion exclusion. We have $k$ choices for the first object, $k-1$ choices for the second, $k-2$ choices for the third, etc... and so on. This gives us $k(k-1)(k-2)\cdots (k-n+1)$ ways to distribute the objects into the boxes. However, this counts the number of ways to distribute the objects without regard to the order in which they are placed into the boxes. Since the order in which the objects are placed does not matter, we must divide by $n!$ to correct for this overcounting. Thus, the number of ways to distribute the objects is
$$\frac{k(k-1)(k-2)\cdots (k-n+1)}{n!} = \frac{k!}{(k-n)!n!}$$
The number of ways to distribute the objects such that each box receives at least one object is then
$$\frac{k!}{(k-n)!n!}-\frac{k!}{(k-n-1)!n!}$$
This simplifies to
$$\frac{k!}{(k-n)!n!}-\frac{k!}{(k-n-1)!n!} = \frac{k!}{(k-n)!n!}\left(1-\frac{n}{k-n-1}\right) = \frac{k!}{(k-n)!n!}\left(\frac{k-n-1}{k-n-1}\right) = \frac{k!}{(k-n)!n!}$$
Thus, the number of ways to distribute $n$ indistinguishable balls into $k$ distinguishable boxes such that each box contains at least one ball is $\boxed{\binom{k}{n}}$.
# Answer
\binom{k}{n}$$
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Sehr schön!!!! Perfekt für jemanden, der wie ich süchtig nach Wanda ist
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Πολύ καλή τιμή και ήρθε γρήγορα
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