r3l3ntl3ss
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I'm stuck on this question:

Each of n people drops his business card into an urn. They each then choose a card at random from the urn. Write down a formula for the probability that none of the n people gets his own card.

The answer is \displaystyle\sum_{j=0}^n \dfrac{(-1)^j}{j!} but I'm not sure how.

Appreciate any help, thanks.
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Betelgeuse-
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Its 50% - Either the N People get their own card or they dont.
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Gregorius
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(Original post by r3l3ntl3ss)
I'm stuck on this question:

Each of n people drops his business card into an urn. They each then choose a card at random from the urn. Write down a formula for the probability that none of the n people gets his own card.

The answer is \displaystyle\sum_{j=0}^n \dfrac{(-1)^j}{j!} but I'm not sure how.

Appreciate any help, thanks.
This is what is called a derangement. Probably the slickest way of doing this is via the inclusion exclusion formula. Do you know about that?
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r3l3ntl3ss
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(Original post by Gregorius)
This is what is called a derangement. Probably the slickest way of doing this is via the inclusion exclusion formula. Do you know about that?
yep I know about the inclusion exclusion formula, but I'm unsure about how to begin
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Gregorius
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(Original post by r3l3ntl3ss)
yep I know about the inclusion exclusion formula, but I'm unsure about how to begin
Let A_i be the event that person i does get their card. You want 1 - \mathbb{P}(A_1 \cup A_2 \cup \cdots \cup A_n). Now start crunching the formula.
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r3l3ntl3ss
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(Original post by Gregorius)
Let A_i be the event that person i does get their card. You want 1 - \mathbb{P}(A_1 \cup A_2 \cup \cdots \cup A_n). Now start crunching the formula.
ah I see, thank you!
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