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  #1  
Old 02-23-2005, 10:13 AM
jskills jskills is offline
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Default proposing a toast - clinking glasses

Off topic somewhat, but I know there is a formula to determine the number of times glasses are clinked together at a table when everyone toasts.

For example:

- if there are only 2 people at the table, there is 1 clink.
- with 3 people, you have 3 clinks
- with 4 people, you have 6 clinks
etc ...


Is it:

(n-1)!

?
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  #2  
Old 02-23-2005, 10:25 AM
pzhon pzhon is offline
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Default Re: proposing a toast - clinking glasses

The number of pairs among n people is called "n choose 2" and it equals n(n-1)/2. More generally, "n choose k" is the number of ways of choosing a subset of size k from a set of n objects, and it equals n!/(k! (n-k)!).
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  #3  
Old 02-23-2005, 11:51 AM
jskills jskills is offline
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Default Re: proposing a toast - clinking glasses

Thank you very much.

This one comes up at almost every family function I'm at and I continue to not be able to recall it ...
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  #4  
Old 02-23-2005, 03:41 PM
AngusThermopyle AngusThermopyle is offline
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Default Re: proposing a toast - clinking glasses

A way to explain it to your friends/family with out just quoting a formula:

N people.
Each person will 'clink' with N - 1 others.
But each 'clink' is experienced twice, once by each person in the toast.
So N people times N -1 'clinks' divided by 2 persons per 'clink' gives you

N*(N-1)/2
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  #5  
Old 02-24-2005, 03:52 AM
ajizzle ajizzle is offline
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Default Re: proposing a toast - clinking glasses

(n)(n-1)/2. I'm sure someone else has given this already, but i'm too busy to read the thread
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