Worrisome Wedding Workout

Firstly, since there is the greatest restriction with Belle, Benjamin and Bernadette, we need to position the bride and her parents. They need to sit in a block of 3, and we can easily see that there are $9$ possibilities. However, each of these positions have 2 possibilities, either Mum-Belle-Dad or Dad-Belle-Mum, hence there are $9 \times 2 = \boxed{18}$ ways to seat the bride and her parents.

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Next, look at the groom. Since there are no more restrictions about direct seating (only relative to one another), we can just imagine the 8 remaining flexible seats by themselves in the same order as it doesn't matter which 3 have been already taken. Now for the groom and parents, we need to choose 3 of these 8, however, for each of these positions, there are 2 possibilities: Mum-Alan-Dad or Dad-Alan-Mum. Therefore, there are $\dbinom83 \times 2 = \boxed{112}$ ways to seat the groom and his parents.

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For the remaining 5 guests, there are $\dfrac{5!}{2! \times 2!} = \boxed{30}$ ways of arranging them. If you're not sure about this, see this Brilliant.org Wikipage .

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Therefore, the total number of ways of accomplishing your task is $18 \times 112 \times 30 = \boxed{60480}$

First, since the bride and her parents need to sit together, we stack their cards on top of each other and move them as one card; so we're now essentially arranging nine cards into nine positions. Next we decide the relative positions of the groom and his parents: the groom has to sit in the middle, so our choices are father-groom-mother or mother-groom-father, i.e. $\mathbf{2}$ possibilities. Then we choose which of the available nine positions to seat them in, we can do that $\mathbf{\binom{9}{3}}$ ways. Now we arrange the remaining six cards into the six remaining positions; since there are two identical Chris's and two identical Drew's, we can do this in $\mathbf{\frac{6!}{2!2!}}$ ways. Finally once "bride stack" has been positioned, we unstack the cards, again, their relative positions can be father-bride-mother or mother-bride-father, so we have $\mathbf{2}$ ways to do that. Thus the total number ways of accomplishing the task is

$2 \cdot \dbinom{9}{3} \cdot \dfrac{6!}{2!2!} \cdot 2 = \boxed{60480}$

Since the bride and her parents must all sit directly beside each other, we can treat these $3$ name tags as $1$ name tag, which essentially gives us a total of $9$ name tags, which can be arranged in $9!$ ways, but with some restrictions:

The bride and her parents can sit in $2$ different ways (Benjamin-Belle-Bernadette or Bernadette-Belle-Benjamin) so we will need to multiply by $2$ .

There are $2$ repeated Chris name tags, so we will need to divide by $2!$ (see here ), and $2$ repeated Drew name tags, so we will need to divide by $2!$ again.

Since the groom must sit somewhere between his parents, we can treat his name tag and his parents' name tags as repeated name tags, so we will need to divide by $3!$ ; but there are $2$ different ways that the groom can sit between his parents (Alexander-Alan-Angelina or Angelina-Alan-Alexander) so we will need to multiply by $2$ again.

Therefore, the number of ways of placing the name tags is then $\frac{9! \cdot 2 \cdot 2}{2! \cdot 2! \cdot 3!} = \boxed{60480}$ .