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#s of combinations

 
 
Levi
 
Reply Thu 14 Oct, 2004 03:20 pm
I understand that to figure out how many different combinations there are when you have, for example, 5 houses and 5 plots of land...

5! = 5*4*3*2*1 = 120 different combinations

But what about when a number differs? What if there are 6 houses and 5 plots of land?
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ebrown p
 
  1  
Reply Thu 14 Oct, 2004 03:25 pm
You need to state the problem better. The way you have it now is ambiguous (it can be understood in more than one way).

Can you put all 6 houses on 1 plot of land?
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Levi
 
  1  
Reply Thu 14 Oct, 2004 06:28 pm
When I posted I meant no more than one house per plot... but now that you mention it I'm interested in the difference in how it's worked out when any number of the six houses can be on one plot... So if you wouldn't mind explaining it both ways. :wink:

Or are there any other uncertainties?
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ebrown p
 
  1  
Reply Thu 14 Oct, 2004 07:44 pm
In that case there will always be exactly one plot with no house. So why not imagine that there is a "no house" (that is a house that doesn't exist). Then you can solve the provlem the same as you did before.

Does this make sense?
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markr
 
  1  
Reply Thu 14 Oct, 2004 07:49 pm
Code:
H = # houses
P = # plots
H >= P

1 house/plot

Houses distinguishable, Plots distinguishable
P(H,P) = H!/(H-P)! (# of permutations of H things taken P at a time)
For H=3, P=2, there are 6 ways:
A B
A C
B A
B C
C A
C B

Houses distinguishable, Plots indistinguishable
C(H,P) = H!/P!(H-P)! (# of combinations of H things taken P at a time)
For H=3, P=2, there are 3 ways:
A B
A C
B C

Houses indistinguishable
1 (each plot gets any house)
For H=3, P=2, there is 1 way:
H H


Any number of houses/plot

Houses distinguishable, Plots distinguishable
P^H (P choices for each house)
For H=3, P=2, there are 8 ways:
ABC {}
AB C
AC B
BC A
A BC
B AC
C AB
{} ABC

Houses distinguishable, Plots indistinguishable
This is the sum of the # of unique arrangements of houses for each of the ways to partition H into at most P partitions.
For H=3, P=2, there are 4 ways:
ABC {}
AB C
AC B
BC A

Houses indistinguishable, Plots distinguishable
C(H+P-1,H) = (H+P-1)!/H!(P-1)!
For H=3, P=2, there are 4 ways:
HHH {}
HH H
H HH
{} HHH

Houses indistinguishable, Plots indistinguishable
This is the number of ways to partition H into at most P partitions.
For H=3, P=2, there are 2 ways:
HHH {}
HH H
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