Modified Derangement : In how many ways can they enter these houses so that no one goes into their own house.

Question

There are four persons $A,B,C$ and $D$ such that $A$ has two houses whilst $B,C$ and $D$ have one house each. In how many ways can they enter these houses so that no one goes into their own house.

My Attempt: Number of Derangement amongst $5$ persons and $5$ houses is given by $D_{5}=44$. But here there are only $4$ persons. So we need to subtract that case in which the fifth person say $E$ goes into one of the $A's$ houses. After this I am not able to make any progress.

Answer 1

Let there be five people $A,B,C,D,E$, each owning only one house. We want the count of derangements in which $A$ doesn't go to $E$'s house.

Among the total $D_5=44$ derangements, $A$ is equally likely to go to $B,C,D,E$'s houses. So $A$ does not go to $E$'s house in $\dfrac{1}{4}D_5$ ways.

Number of ways $= D_5 - \dfrac{1}{4}D_5 = 33$

It is okay for $E$ to go to $A$'s house; it means only one of $A$'s two houses is occupied.

But to this, we need to add the case when $E$ goes to his own house and others go to different houses, since this is allowed. This is just $D_4=9$.

Hence final answer is $\boxed{33+9=42}$.

Answer 2

Let's say $A$ has a red house and a blue one. If no one goes to the red house then we just get a usual derrangement so there is $9$ options.

If someone goes to the red house select a person $x$ that goes to the red house (there is $3$ options they are all the same), there is now two cases:

1. Select a person $y$ that goes to $x$'s house ($3$ ways) and now note that there are $3$ options for the remaining two vertices (one case is when no one goes to $y$'s house and the other $2$ ways are when one of them goes to $y$'s house), so $9$ cases total.

2. if no one goes to $x$'s house, draw the graph for this case and note it is just another derrangement with $3$ vertices so we get $2$ cases.

Hence the answer is $9 + 3(2 + 9) = 42$.

Answer 3

Lets think using a simple way such as P.I.E:

-We know that all placements without restriction is $P(5,4)=120$

Now , we must subtract the cases where $A$ or $B$ or $C$ or $D$ are their own houses such that $$|A|+|B|+|C|+|D|-[|A \cap B|+|A \cap C|+ |A \cap D|+|B \cap C| + |B \cap D| + |C \cap D|] + [|A \cap B \cap C|+|A \cap B \cap D|]+|A \cap C \cap D|+|B \cap C \cap D| - |A \cap B \cap C \cap D|$$ where

$|A|=2 \times 4! , |B|=4! ,|C|=4! , |D|=4! ,|A \cap B|=|A \cap C|= |A \cap D|=2 \times 3!,|B \cap C| = |B \cap D| = |C \cap D|=3! ,|A \cap B \cap C|=|A \cap B \cap D|=|A \cap C \cap D|=4 ,|B \cap C \cap D|=2 ,|A \cap B \cap C \cap D|=2 $

Then $$P(5,4) -(5 \times 4!)+(9 \times 3!) -(14)+2=42$$