Simple but tricky permutation question There are $72$ perspective home-buyers, and $300$ houses in a line. However, assume only sixty of the people purchase a house. Among the people are Oscar and Patricia, they were divorced recently, and refuse to ever be neighbors.  How many possible housing arrangements are there?
Remember that neither Patricia nor Oscar are necessarily part of the $60$.
I reached the answer of: $_{72}C_{60} \cdot \mathbb{P}(300,60)-2\left(_{70}C_{58}\right)\cdot\mathbb{P}(299,59)$, but I don't think I'm correct...
 A: Total number of arrangements: $\binom{72}{60} \binom{300}{60} 60!$. (This is the same as what you computed.)
Arrangements where Oscar and Patricia are neighbors: $\binom{70}{58} 299 \cdot 2 \cdot \binom{298}{58} 58!$.

*

*$\binom{70}{58}$ ways to choose a set of $58$ buyers besides Oscar and Patricia

*$299$ ways to choose a pair of adjacent houses for Oscar and Patricia

*$2$ ways to place Oscar and Patricia in the adjacent pair of houses

*$\binom{298}{58}$ ways to choose 58 houses for the remaining 58 buyers

*$58!$ ways to place the 58 buyers in the 58 bought houses.

A: Your result was correct. You can confirm this by algebraically manipulating either your expression or one of the other answers until you are able to show they are equal.
Here's a derivation of the formula corresponding to the form in which you wrote it.
First consider how many ways $60$ of the $72$ buyers could buy houses if there were no restriction against Oscar and Patricia buying adjacent homes.
The set of buyers can be selected in $\binom{72}{60}$ ways,
and for each set of buyers the $60$ houses they buy can be selected in
$P^{300}_{60}$ ways (permutations of $60$ houses out of $300$, since order matters).
So there would be a total of
$$ \binom{72}{60} {P^{300}_{60}} $$
distinct arrangements.
Knowing that Oscar and Patricia will not buy adjacent houses, all of the
$ \binom{72}{60} {P^{300}_{60}} $ arrangements
we have found are valid except the ones in which Oscar and Patricia are adjacent.
So we must subtract the adjacent cases.
These occur only in the cases where both Oscar and Patricia are buyers.
So we only have to be concerned with the
$\binom{70}{58}$ subsets of buyers who include both Oscar and Patricia.
Now suppose we have a selection of $60$ buyers including Oscar and Patricia.
Take the numbers $1$ to $299$ and assign Oscar and Patricia (together)
to one of the numbers and the other $58$ buyers each to a unique number from among the remaining numbers.
If $n$ is the number Oscar and Patricia are assigned to,
Oscar and Patricia buy houses $n$ and $n+1$;
any other buyer assigned to a number $k < n$ buys house $k$,
and any other buyer assigned to a number $k > n$ buys house $k + 1.$
All of this so far can be done in $\binom{70}{58} P^{299}_{59}$ ways,
since for each of $\binom{70}{58}$ subsets
we are assigning each of $59$ identifiable "buying units" to
$299$ distinct integers.
But for each such assignment we can put Oscar in house $n$ and Patricia in house $n+1$ or vice versa, so the total number of possible assignments
including the choice of which houses Oscar and Patricia buy is
$$ 2 \binom{70}{58} P^{299}_{59}. $$
With a little effort you can confirm that every such assignment gives an arrangement of house purchases in which Oscar and Patricia buy adjacent houses,
and every arrangement in which Oscar and Patricia buy adjacent houses is generated by exactly one of these assignments.
So the total number of possible arrangements of house purchases in which
Oscar and Patricia are not neighbors is
$$ \binom{72}{60} {P^{300}_{60}} - 2 \binom{70}{58} P^{299}_{59}, $$
exactly as proposed in the question.
