yougizmos
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Jane, Mary and Oscar are having dinner. Jane makes an eloquent toast about friendship, after which they all "clank" their wine glasses. Jane clanks with Mary, Jane clanks with Oscar, Oscar clanks with Mary - there are 3 clanks in all. The next day, Jane, Mary and Oscar are joined by 5 other friends. If someone makes another toast, how many clanks will there be, assuming that Jane clanking with Mary and Mary clanking with Jane, counts as one clank. What is the algorithm that can be used to figure out how many clanks are produced by (n) number of dinner guest? How many clanks for a toast at a table of 27?
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lancewrath
77 Creations.
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no, don't you know the number that solves all mathmatical equation and even equates the meaning of life is 42!
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lancewrath
77 Creations.
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ok my actual answer for this is 351
my equation is
=N*((N-1)/2)
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drphonon
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This is a case of using the N choose M formula. Given a set of N things how many distinct sets of M things are possible. In this case we have 8 people and want to know how many different sets of 2 people are possible.
N = 8
M = 2
The formula is N!/(M!*(N-M)!).
Here x! means x factorial which is x*(x-1)*(x-2)*...*1
For example 3! = 3*2*1 = 6
4! = 4*3*2*1 = 24
So 8 choose 2 = 8!/(2!*(8-2)!) = 8*7*6*5*4*3*2*1/(2*1*(6*5*4*3*2*1)) = 8*7/2 = 28.
There are 28 clinks.
The first part of the problem is 3 choose 2 = 3!/(2!*(3-2)!)) = 3*2*1/(2*1*1) = 3
With 27 people we have 27 choose 2 = 27!/(2!*25!) = 27*26/2 = 351 clinks.
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lancewrath
77 Creations.
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ah yes, you broke it down alot better than i did.
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m084410
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my answer is also (N*N-1)/2
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Inspire change
