Counting ways to line up for a family photo.

Question

Question: A family lines up for a photograph. In each of the following situations, how many ways are there for the family to line up so that the mother is next to at least one of her daughters?

The family consists of two parents, three daughters and four sons.

So I tried to solve this using the Inclusion-Exclusion principle but got stuck at the very end step. The formula I used is: |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|

Ways for mom to be next to first daughter + Ways for mom to be next to second daughter + Ways for mom to be next to third daughter - Ways for mom to be next to first and second daughter - Ways for mom to be next to first and third daughter - Ways for mom to be next to second and third daughter + Ways for mom to the next to all 3 daughters

I was able to calculate everything except for |A ∩ B ∩ C|. How do we arrange the mom and 3 daughters in this case? Thank you for your help.

Edit: I have $(2⋅8!)$ ways for the mom to be next to one daughter. There are $3$ daughters so we have $3⋅2⋅8!$ ways. And then $(2⋅7!)$ ways for each time a mom is between any 2 daughters, that makes $(3⋅2⋅7!)$ in total. Looks like there's $0$ ways for the mom to be next to all $3$ daughters, so I suppose the answer is $(3⋅2⋅8!) - (3⋅2⋅7!)$.

Answer 1

Maybe it is easier to construct the ways of arranging where no daughter is beside the mother:

• All arrangements of $2$ parents, $3$ daughters, $4$ sons: $\color{blue}{9!}$
• First line up parents and sons: $(2+4)! = \color{blue}{6!}$
• Each arrangement of these $6$ people produce $7$ possible "slots" where to place the daughters: $\color{green}{7}$
• The $\color{green}{2}$ slots left and right to the mother are not allowed: $\Rightarrow \color{green}{7-2} = \color{green}{5}$ slot
• Distribute $\color{blue}{3}$ daughters (permutations not considered) across the "allowed" $\color{green}{5}$ slots $s_1,\ldots , s_5$: $s_1+ \cdots + s_5 = 3$: $\binom{\color{green}{5}+\color{blue}{3}-1}{\color{blue}{3}} = \color{blue}{\binom 73}$ (this can be seen as combination with repetition)
• The $3$ daughters can be arranged among themselves: $\color{blue}{3!}$

All together

$$\color{blue}{9! - 6! \cdot \binom 73 \cdot 3!}$$