7 different cars distributed to three daughters and two sons with restrictions Jay has seven different cars that he is leaving to his three daughters and
two sons. The Maserati must go to a daughter, and the Bentley must go to a son. Each heir is to receive at least one and no more than two cars. How many ways can the cars be distributed?
I have seen similar question but still not sure how to calculate the above question. Any help appreciated. Thanks!
In how many ways can six different gifts be given to five different children with each child receiving at least one gift?
 A: Brute force says if the sons get $2$ cars, choose $1$ of the $5$ brand-X cars to accompany the Bently, choose $1$ of $2$ to give to the first son, choose $2$ of the $3$ daughters to get $2$ cars, choose $2$ of the $5$ daughter cars to give to the first $2$-car daughter and choose $2$ of the remaining $3$ cars to give to the other $2$-car daughter: $900$ ways.  
If the sons get $3$ cars, choose $2$ of $5$ brand-X for them, choose $1$ of $2$ sons to get $2$ cars, choose $2$ of $3$ cars to give to the $2$-car son, choose $1$ of $3$ daughters to get $2$ cars, choose $2$ of $4$ cars to give to that daughter, and then $1$ of $2$ to give to the first $1$-car daughter: $2160$ ways.  
It the sons get $4$ cars, choose $3$ of $5$ brand-X to give to sons, choose $2$ of $4$ to give to the first son, choose $1$ of $3$ to give to the first daughter and choose $1$ of $2$ to give to the second: 360 ways.  
So the answer really is $3420$ ways.  
EDIT: For a more formulaic approach to the problem, suppose we count the number of ways $m$ objects can be placed in $j$ indistinguishable boxes, each box getting $1$ or $2$ objects. There must be $m-j$ boxes that get $2$ objects and $2j-m$ boxes that get $1$ object. Put the objects in slots where the slots for the $2$-object boxes come first and slots belonging to the same box are consecutive. An example for $m=8$, $j=5$:
$$\begin{array}{r|cc|cc|cc|c|c|}
\text{slot}&1&2&3&4&5&6&7&8\\
\text{object}&7&8&1&6&4&2&3&5\end{array}$$
There are $m!$ ways to put the objects in the slots, but for each $2$-object box there are $2!=2$ placements into the box which are equivalent. Also there are $(m-j)!$ ways the $m-j$ $2$-object boxes could be given the same contents but in different order and $(2j-m)!$ ways the $1$-object boxes could be given the same contents but in different order. Thus there are
$$\frac{m!}{2^{m-j}(m-j)!(2j-m)!}$$
ways to put $m$ objects in $j$ indistinguishable boxes, each box getting $1$ or $2$ objects. Having said as much, if the boxes were distinguishable there would be $j!$ ways to label them for each distribution, so the number of ways to put $m$ distinguishable objects in $j$ distinguishable boxes is
$$\frac{m!j!}{2^{m-j}(m-j)!(2j-m)!}$$
So for our problem the boys could get $k\in\{1,2,3\}$ brand-X cars along with their Bently so for them $m=k+1$ and $j=2$. The girls get the Maserati and the other $5-k$ brand-X cars so for them $m=6-k$ and $j=3$. There are $5\choose k$ ways to select brand-X car to give the boys, the girls getting the rest. The number of ways the cars can be distributed is therefore
$$\sum_{k=1}^3{5\choose k}\frac{(k+1)!2!(6-k)!3!}{2^{k-1}(k-1)!(3-k)!2^{3-k}(3-k)!k!}=900+2160+360=3420$$
ways as already worked out but prehaps less systematically.
A: The only way I can think of requires a case-by-case analysis.  First, there are $3\times 2 = 6$ ways to distribute the Maserati and the Bentley.  How to distribute the remaining $5$ cars?
Case A: Ms. Maserati got another car and Mr. Bentley got another car: Just line up the $5$ kids in birth order and give them $1$ car each.  No. of ways $= 5!$
Case B: Ms. Maserati got another car but not Mr. Bentley: Line up the $4$ kids (everyone except Mr. Bentley) in birth order, pick one of the $3$ non-Maserati kids (lets call him/her Pat) to have two cars, and then distribute the $5$ cars.  No. of ways $=3\times 5! /2! $ where the division by $2!$ accounts for the fact Pat has two cars.
Case C: Mr. Bentley got another car but not Ms. Maserati: Same math as Case B.
Case D: neither Ms Maserati nor Mr. Bentley got another car: Line up the $3$ other kids in birth order, pick one of the $3$ to have a single car, then distribute the $5$ cars.  No. of ways $= 3 \times 5 ! / 4$ where the division by $4=2!\times 2!$ accounts for the fact that two kids have two cars each.
Total $= 6 \times 5! \times (1 + \frac32 + \frac32 + \frac34) = 3420$
Note: the dividing by $2$ in some cases and by $4$ in some other cases is the "factoring out duplication conditional on a case analysis" I mentioned in the comments.
