In how many ways can we seat 3 pairs of siblings in a row of 7 chairs,
so that nobody sits next to his or her sibling? [One chair will be empty]
I got that there are $7!$ ways of seating everyone with no restrictions. There are $2 \cdot 6!$ ways for each pair of siblings to sit together, and since there are $3$ siblings, there are $2 \cdot 3 \cdot 6!$ for at least one pair siblings to sit together.
Thus, I got that there are $7! – 6 \cdot 6!$ ways in which nobody sits next to his or her siblings.
However, it seems that this answer is incorrect and you must add back the number of seatings with two pairs sitting together, then subtract off the number of ways to have every pair of siblings sitting together.
What is the reason for doing this? I don't see why you need to add back the number of seatings with two of the pairs sitting together.
Best Answer
Suppose we have six persons $$a_1,a_2,b_1,b_2,c_1,c_2$$ and a row of seven chairs. Consider $U$ as the set of all ways of seating everyone with no restrictions. Consider $A$ as the set of all ways of seating s.t. $a_1$ and $a_2$ are adjacent. Consider the set $B$ in the same manner for $b_i$'s and the set $C$ for $c_i$'s. In this question we are looking for the size of the set $$A^c\cap B^c\cap C^c$$ but $$\begin{align*} |A^c\cap B^c\cap C^c|& =|(A\cup B\cup C)^c| \\ &= |U|-|A\cup B\cup C| \end{align*}$$ and $$\begin{align*} |A\cup B\cup C|=& |A|+|B|+|C| \\ &-|A\cap B|-|A\cap C| -|B\cap C| \\ & +|A\cap B\cap C| \\ =& 3\times 2\cdot6! \\ &-3\times2\cdot 2\cdot 5! \\ &+2\cdot 2\cdot2\cdot 4! \end{align*}$$ so the final answer is $$7!-6\cdot6!+2\cdot6!-8\cdot4!=3\cdot6!-8\cdot4!=82\cdot4!=82\cdot24=1968.$$
OK, I see what's the problem. You use the equality $$|A\cup B\cup C|= |A|+|B|+|C|,$$ but you have to note that this equality holds only if $$A\cap B=A\cap C=B\cap C=\emptyset,$$ and that my friend is not true here; these intersections in our specific problem are not empty, and that makes the problem a little complicated because for example when you add $|A|$ and $|B|$ you count the size of the intersection 2 times once in $|A|$ and once in $|B|$, so it's reasonable to subtract it once, and that's the whole idea behind "Inclusion Exclusion Principle": you have to "Include" each proper case only once, and extra counting should be "Exclude". I hope it helps you to understand the idea better.