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Forum > Africa AAA League > Africa AAA #1 > let's brush up on enumerative combinatorics
Catullus16
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starting with the basics.

like combinatorial analysis and axioms of probability, stuff like that.
 
Catullus16
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The game of bridge is played by 4 players, each of
whom is dealt 13 cards. How many bridge deals are
possible?
 
Catullus16
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Originally posted by Catullus16
The game of bridge is played by 4 players, each of
whom is dealt 13 cards. How many bridge deals are
possible?


so 52! unique permutations, but card order doesn't matter within any given hand, so 52!/(13!13!13!13!) or 52!/(13!^4) which is like 53 octillion or something
 
Catullus16
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53644737765488792839237440000

dang, son
 
Catullus16
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okay, binomial and multinomial expansion, how do they work again
 
Catullus16
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Originally posted by Catullus16
okay, binomial and multinomial expansion, how do they work again


https://en.wikipedia.org/wiki/Multinomial_theorem
https://www.youtube.com/watch?v=cJsNNcQlSPA
 
Catullus16
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wikipedia is kinda the best math textbook ever

and mathdoctorbob is a lovable musclenerd of a hero:
https://www.youtube.com/watch?v=MVmgsATTg2I

also patrickJMT: https://www.youtube.com/watch?v=1pSD8cYyqUo

so basically (a+b)^n => sum from k=0 to n of the quantity [ nCk * a^k * b^(n-k) ]

okay, that works. just need to know how to use choose functions to efficiently check coefficients.
 
Catullus16
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If 8 new teachers are to be divided among 4
schools, how many divisions are possible? What if
each school must receive 2 teachers?
 
Catullus16
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Originally posted by Catullus16
If 8 new teachers are to be divided among 4
schools, how many divisions are possible? What if
each school must receive 2 teachers?


second part is easier -- just 8!/(2!^4)=2520

first part is reversing the multinomial sum. so 8 choose a,b,c,d where a+b+c+d=4, so reversing the sum yields the polynomial (1+1+1+1)^8 = 4^8 = 65536
 
Catullus16
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Ten weight lifters are competing in a team weightlifting contest. Of the lifters, 3 are from the United
States, 4 are from Russia, 2 are from China, and 1
is from Canada. If the scoring takes account of the
countries that the lifters represent, but not their
individual identities, how many different outcomes
are possible from the point of view of scores? How
many different outcomes correspond to results in
which the United States has 1 competitor in the
top three and 2 in the bottom three?
 
Catullus16
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Originally posted by Catullus16
Ten weight lifters are competing in a team weightlifting contest. Of the lifters, 3 are from the United
States, 4 are from Russia, 2 are from China, and 1
is from Canada. If the scoring takes account of the
countries that the lifters represent, but not their
individual identities, how many different outcomes
are possible from the point of view of scores? How
many different outcomes correspond to results in
which the United States has 1 competitor in the
top three and 2 in the bottom three?


this is like the earlier bridge hands question where it's basically a permutation of combinations
10 lifters total, so numerator is 10!
and then 3!4!2!1! for the demonator
so 10 choose 1,2,3,4 = 12600

for the second question, there are (3c1)(3c2)=9 ways to have american lifters place that way, so take the product of that and the permutative combinations of the other lifters, so 7 choose 1,2,4 = 105
and 9*105 = 945
 
Catullus16
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Delegates from 10 countries, including Russia,
France, England, and the United States, are to
be seated in a row. How many different seating arrangements are possible if the French and
English delegates are to be seated next to each
other and the Russian and U.S. delegates are not
to be next to each other?
 
Catullus16
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Originally posted by Catullus16
Delegates from 10 countries, including Russia,
France, England, and the United States, are to
be seated in a row. How many different seating arrangements are possible if the French and
English delegates are to be seated next to each
other and the Russian and U.S. delegates are not
to be next to each other?


treat the F+E as a single block, so the overcount is double 9!

and then subtract the ways R+US could be a block, so the adjustment is 2!8!

so we have 2(9!-2!8!) = 2(9!)-4(8!) = 564480
 
damf
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Originally posted by Catullus16
53644737765488792839237440000

dang, son


53644737765488792839237440001
 
Catullus16
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always gotta one-up me, well played
 


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