Quick Combinatorics Question (1 Viewer)

[amdspotter]

Its a simple question but I'm having a mindblank, if anyone could quickly explain how to answer this combinatorics question it would be appreciated. I have attached an image of the question below

 

[quickoats]

This is probably easiest to think of it in a not-so-intuitive way. Instead of the heirs picking the houses, the houses pick the heirs i.e.

House 1 has a "choice" of being distributed to A, B or C
House 2 has a "choice" of being distributed to A, B or C
...
House 10 has a "choice" of being distributed to A, B or C

So we can see there are 3^10 ways of 10 houses being distributed between 3 people. NOTE that this includes the case where some people may not get any houses at all (this can happen with inheritances when people squabble). If everyone must have at least 1 house, then you must subtract the cases where one or two people get no houses at all.

 

[amdspotter]

@quickoats
Oh ok, but i still dont quite understand the logic behind the idea of the houses picking their heirs. Like is their a specific reason as to why it must be done like that and was there anything in the question that made you do it in that way? Thanks a lot btw

 

[quickoats]

Easier example that might solidify things. 10 different books to be divided between 2 people. A person is allowed to have no books.

Lets line up the books in a row (on a table perhaps):

B1 B2 B3 B4 B5 B6 B7 B8 B9 B10

They can either go to person A or person B. Let's just put person A on top of the table and person B below the table like so:




Person A

B1 B2 B3 B4 B5 B6 B7 B8 B9 B10

Person B




Each book can go "up" or "down" to get sorted to a respective person i.e.

Person A

Person B

Each book has 2 options, so we have the total cases = 2 x 2 x 2 ... x 2 (10 times) = 2^10. This is an almost identical reasoning to the houses qn.

Why do we use this method?
It is much more difficult to break the 10 houses up into different cases i.e.
Case X: A gets 1 house, B gets 3 houses, C gets 6 houses.
Total = 10C1 x 9C3 x 6C6
Case Y: A gets 2 houses, B gets 5 houses, C gets 3 houses.
Total = 10C2 x 9C5 x 3C3
There are many many (many) more cases to consider and this is too much to do....

 

[amdspotter]

@quickoats thanks a lot I understand the question now really appreciate it