MHB Probability of Male Picking Wife as Dance Partner

[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 pick his own wife as his dancing partner.

 

[soroban]

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.

Spoiler

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

 

[anemone]

Hello, anemone!

Spoiler

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.