MHB Probability of Male Picking Wife as Dance Partner

AI Thread Summary
The discussion revolves around calculating the probability that at least one male at a party of six married couples chooses his own wife as a dance partner. Participants share their methods for solving the problem, with one user expressing satisfaction at discovering a more efficient approach. The complexity of the problem is acknowledged, indicating that it can be challenging to arrive at the correct solution. The conversation highlights the collaborative nature of problem-solving in mathematical discussions. Overall, the focus remains on the probability calculation related to the dance partner selection.
anemone
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There are 6 married couples (12 people) in a party. If every male has to pick a female as his dancing partner, find the probability that at least one male pick his own wife as his dancing partner.
 
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Hello, anemone!

There are 6 married couples (12 people) in a party.
If every male has to pick a female as his dancing partner,
find the probability that at least one male picks his own wife
as his dancing partner.
There are $6! = 720$ possible outcomes.

If no man picks his own wife,
$\quad$ this is a derangement of 6 objects.
$d(6) \,=\,265$

$P(\text{no married couples}) \:=\:\dfrac{265}{720} \:=\:\dfrac{53}{144}$

Therefore: $\:P(\text{some married couples}) \;=\;1 - \dfrac{53}{144} \;=\;\dfrac{91}{144}$
 
soroban said:
Hello, anemone!


There are $6! = 720$ possible outcomes.

If no man picks his own wife,
$\quad$ this is a derangement of 6 objects.
$d(6) \,=\,265$

$P(\text{no married couples}) \:=\:\dfrac{265}{720} \:=\:\dfrac{53}{144}$

Therefore: $\:P(\text{some married couples}) \;=\;1 - \dfrac{53}{144} \;=\;\dfrac{91}{144}$

Well done, soroban! I solved this problem the very hard way, thus I'm happy to see this other approach to say the least.:o
 
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