Math about Christmas Meals (Corrected)

This question can be easily solved using trial and error (due to the small cases) but I have used a slightly different technique known as the Rook's Polynomial. (If you are interested, I have posted an introduction on it here )

First, construct a $4 \times 4$ board $B_{4,4}$ as seen in the diagram. Label the rows with the names of each children (the first letter of the names as a short form), and the columns with the names of the set meals. Let the restrictions (e.g. Carrie doesn't eat chicken) be shaded in green.

Diagram

Then, lets reconstruct $B_{4,4}$ into disjoint subboards by switching row(s) or/and column(s) (in this case, done by switching rows $B$ and $C$ .) From the diagram below (with blue shaded squares), we can see that there are $3$ disjoint subboards ( $B_1, B_2, B_3$ ).

Diagram 2

Here, using Rook Polynomial (do look at the intro), we get

$R(x, B_{4,4}) = R(x, B_1)R(x, B_2)R(x, B_3) = (x^2 + 3x+1)(x+1)^2 = x^4 + 5x^3 +8x^2 +5x +1$

Then, using the inclusion-exclusion formula, our answer is

$4! - 5(3!) + 8(2!) - 5(1!) +1(0!) = \boxed{6}$ .

How I solved it was I represented each person as a letter ( $A$ for Adam, $B$ for Ben, $C$ for Carrie, and $D$ for David. Then I represented the meals as subscript numbers ( $1$ for Chicken Burger, $2$ for Vegetable Frenzy, $3$ for Spicy Gala, and $4$ for Steak. After reading the last paragraph, you get the following:

$A_{34} - B_{124} - C_{234} - D_{123}$

Now this is really easy because, since the children don't want to be copycats, when you assign a particular meal ( $1$ , $2$ , $3$ , or $4$ ), you can be sure that nobody else will get it, therefore allowing you to 'cross it off the list'. In general, this is just a basic combinatorics problem.

Okay so all we want to do is see how many ways we can organize the numbers $1$ , $2$ , $3$ , and $4$ , but with the restrictions/conditions as stated above (make sure no kid is assigned a meal they don't want and no copycats).

$A_{3} - B_{1} - C_{4} - D_{2}$

$A_{3} - B_{2} - C_{4} - D_{1}$

$A_{3} - B_{4} - C_{2} - D_{1}$

$A_{4} - B_{1} - C_{2} - D_{3}$

$A_{4} - B_{1} - C_{3} - D_{2}$

$A_{4} - B_{2} - C_{3} - D_{1}$

That's a total of $\boxed{6}$ combinations!

Let's abbreviate our meals as CB, VF, SG, and S. We know that Adam can't eat CB or VF, so he can only eat SG. Ben can only eat CB, VF, and S; Carrie can only eat VF, SG, and S; and David can only eat CB, VF, and SG. Since there are 24 total arrangements (they don't want to be "copycats") that may or may not satisfy their wants, we can just systematically list out the possible ways that work.

Adam----------Ben----------Carrie---------David

SG................CB............S..................VF

SG................VF............S..................CB

SG................S...............VF...............CB

S...................CB............VF...............SG

S...................CB............SG...............VF

S...................VF............SG...............CB