Find the probability that there is no king and queen of the same suit next to each other.
My attempt: Let $A_{1}$ denote where $K\heartsuit\,Q\heartsuit$ occurs, $A_{2}$ denote where $K\clubsuit\,Q\clubsuit$ occurs, $A_{3}$ denote where $K\diamondsuit\,Q\diamondsuit$ occurs, $A_{4}$ denote where $K\spadesuit\,Q\spadesuit$ occurs,
Then $\lvert A_{1}\cup A_{2}\cup A_{3}\cup A_{4}\rvert = 4(7!)-6(6!)+4(5!)-4!.$ Let $U$ be the set of all the Kings and Queens in the standard deck. So, $|U|=8!$. Then $$|U|-|A_{1}\cup A_{2}\cup A_{3}\cup A_{4}|=8!-[4(7!)-6(6!)+4(5!)-4!]=24024$$ Therefore the probability is $$\frac{24024}{\binom{52}{8}}$$
I'm not sure if I am doing this problem correctly. Inclusion/exclusion always threw me for a loop. Please help me understand how to do this problem.
If there were no restrictions, we could arrange the eight cards in the set $$\{Q\clubsuit, K\clubsuit, \color{red}{Q\diamondsuit}, \color{red}{K\diamondsuit},\color{red}{Q\heartsuit}, \color{red}{K\heartsuit}, Q\spadesuit, K\spadesuit\}$$ in $8!$ ways.
We must exclude those arrangements in which there are one or more suits in which cards of the same suit are adjacent.
One suit in which the cards are adjacent: Suppose the two clubs are adjacent. Place them in a box marked clubs. We then have seven objects to arrange, the box and the other six cards. They can be arranged in $7!$ ways. The $Q\clubsuit$ and $K\clubsuit$ can be arranged within the box in $2!$ ways. Hence, the number of arrangements in which $Q\clubsuit$ and $K\clubsuit$ are adjacent is $7!2!$. By symmetry, there are an equal number of arrangements in which two hearts, two diamonds, or two spades are adjacent. Thus, there are $$\binom{4}{1}7!2!$$ arrangements in which there is a suit in which two cards of the same suit are adjacent.
However, if we subtract $\binom{4}{1}7!2! = 8!$ from the total, we will have subtracted those cases in which there is more than one suit in which the cards are adjacent more than once.
Two suits in which cards of the same suit are adjacent: Suppose both the two clubs are adjacent and the two diamonds are adjacent. Place the two clubs in a box marked clubs and the two diamonds in a box marked diamonds. We then have six objects to arrange, the two boxes and the other four cards. They can be arranged in $6!$ ways. Within each box, the cards can be arranged in $2!$ ways. Hence, there are $6! \cdot (2!)^2$ arrangements in which the two clubs are adjacent and the two diamonds are adjacent. By symmetry, there are an equal number of arrangements for any pairing of suits. Hence, there are $$\binom{4}{2}6!(2!)^2$$ arrangements in which there are two suits in which the cards of the same suit are adjacent.
Three suits in which cards of the same suit are adjacent: Suppose the two clubs are adjacent, the two diamonds are adjacent, and the two hearts are adjacent. Place the clubs in a box marked clubs, the diamonds in a box marked diamonds, and the hearts in a box marked hearts. We have five objects to arrange, the three boxes and the other two cards. They can be arranged in $5!$ ways. The cards within each box can be arranged in $2!$ ways. Hence, there are $5!(2!)^3$ arrangements in which two clubs are adjacent, two diamonds are adjacent, and two hearts are adjacent. By symmetry, there are an equal number of arrangements for any trio of suits. Hence, there are $$\binom{4}{3}5!(2!)^3$$
arrangements in which there are three suits in which cards of the same suit are adjacent.
Four suits in which cards of the same suit are adjacent: We place the clubs in a box marked clubs, the diamonds in a box marked diamonds, the hearts in a box marked hearts, and the spades in a box marked spades. We have four objects to arrange. Within each box, we can arrange the cards in $2!$ orders. The number of such arrangements is $$\binom{4}{4}4!(2!)^4$$
By the Inclusion-Exclusion Principle, the number of permissible arrangements is $$8! - \binom{4}{1}7!2! + \binom{4}{2}6!(2!)^2 - \binom{4}{3}5!(2!)^3 + \binom{4}{4}4!(2!)^4$$
Thus, the desired probability is $$\frac{8! - \binom{4}{1}7!2! + \binom{4}{2}6!(2!)^2 - \binom{4}{3}5!(2!)^3 + \binom{4}{4}4!(2!)^4}{8!}$$