Probability of 'top six' match-ups on a Premier League match-day There are 20 teams in the English Premier League. Of these 20, there is generally considered to be a clearly defined 'top six' - Arsenal, Manchester United, Manchester City, Liverpool, Chelsea and Tottenham.
Let us assume that on a given match-day, these twenty teams are paired randomly. 
I am trying to find the following on a given match-day:
$P(a)$ is the probability that each 'top six' team plays another 'top six' team
$P(b)$ is the probability that there are exactly two match-ups between 'top six' teams
$P(c)$ is the probability there is exactly one match-up between 'top six' teams
I know that the number of ways to split a group of $n$ elements into pairs is given by:
$$X(n)= \frac{n!}{2^{\frac{n}{2}}(\frac{n}{2})!}$$
So, to find $P(a)$, there are $X(6)$ ways to pair off the 'top six' teams and for each of those ways, there are $X(14)$ ways to pair off the remaining teams. Thus, 
$$P(a)=\frac{X(6)X(14)}{X(20)} = 0.0031$$
To find $P(b)$, my plan was to find the probability of at least two match-ups and then subtract $P(a)$ from this. This is where I get a bit uncertain. There are $\binom{6}{4}$ ways to select groups of 4 'top six' teams and then 3 different ways to create two pairs from any given group of 4. And then for each of set of two pairs, there are $X(16)$ ways to pair off the remaining teams. Thus,
$$P(b)=\frac{\binom{6}{4}X(4)X(16)}{X(20)}-P(a)= 0.136 $$ 
Finally, there are $\binom{6}{2}$ ways to select a pair of 'top six' teams. And for each pair, there are $X(18)$ ways to pair off the remaining teams. 
$$P(c)=\frac{\binom{6}{2}X(18)}{X(20)}-P(a)- P(b)= 0.650 $$ 
Adding these together, there should be a 78.9% chance of at least 1 'top six' match-up occurring on any given match day. 
However, if I calculate the odds at least one 'top six' match-up, wouldn't it also just be:
$$1 - (\frac{14}{19})(\frac{13}{17})(\frac{12}{15})(\frac{11}{13})(\frac{10}{11}) = 0.653$$
Where am I going wrong? Is there a different way that I should be thinking about this? 
 A: For $P(a)$ you have $\frac 5{19}$ that Arsenal is paired with another top six team, then $\frac 3{17}$ that the next one left on your list is paired with a top six team, and finally $\frac 1{15}$ that the next one left is paired with the last team.  $$P(a)=\frac {5\cdot 3 \cdot 1}{19\cdot 17 \cdot 15}=\frac 1{323}$$  
For $P(c)$ you can select the two teams to be matched in $15$ ways.  Having done that, the first one has $\frac 1{19}$ chance of getting the proper opponent, then each of the other four must find a non-top six opponent.  You did not make sure the four teams that should not be paired with a top six opponent in fact are not.
$$P(c)=15\cdot \frac 1{19}\cdot \frac {14}{17}\cdot \frac {13}{15}\cdot \frac {12}{13}=\frac{168}{323}\approx 0.520$$  
For $P(b)$ as you say there are ${6 \choose 4}$ ways to choose the four paired teams and $3$ ways to match them up.  Given a required matching we have $\frac 1{19}$ of getting the first match, $\frac 1{17}$ of getting the second, and $\frac {14}{15}$ of avoiding the match of the last two.  $$P(b)=45\cdot \frac 1{19}\cdot \frac 1{17}\cdot \frac {14}{15}=\frac {42}{323}\approx 0.130$$
A: Here is an approach using the Principle of Inclusion / Exclusion (PIE) and a generating function.  The resulting solution has some similarity to the approach outlined in the OP, but we use PIE to avoid over-counting.
Using notation in the OP, there are $X(20)$ possible matchings of the $20$ teams, all of which we assume are equally likely.  Let's say the Top Six are the teams numbered $1$ through $6$, and say a matching of the teams has "Property $(i,j)$" if Top Six teams $i$ and $j$ are matched, for $1 \le i < j \le 6$. For the purpose of applying PIE, we define $S_i$ as the total of the probabilities of the arrangements that have $i$ of the properties, for $i=1,2,3$.
For $S_1$, we have one pair of Top Six teams which are matched.  There are $\binom{6}{2}$ ways to pick the two teams.  Then these teams can be matched in $X(2)$ ways, and the remaining teams can be matched in $X(18)$ ways.  Since there are $X(20)$ equally likely arrangements in all,
$$S_1 = \binom{6}{2} \frac{X(2) X(18)}{ X(20)}$$
Similarly,
$$\begin{align}
S_2 &= \binom{6}{4} \frac{X(4) X(16)}{X(20)} \\
S_3 &= \binom{6}{6} \frac{X(6) X(14)}{X(20)} \\
\end{align}$$
We define the generating polynomial of $S_i$ as
$$f(x) = 1 + S_1 x + S_2 x^2 + S_3 x^3$$
so supplying numerical values for the $S_i$'s, 
$$f(x) = 1 +0.789474 x+0.139319 x^2+0.00309598 x^3$$
In this form, $f(x)$ is not very interesting.  But by PIE, $f(x-1)$ is the generating function of the arrangements that have exactly $n$ of the properties, i.e. the coefficient of $x^n$ in $f(x-1)$ is the probability of exactly $n$ matches among the Top Six teams.  After substituting $x-1$ for $x$ in $f(x)$ and expanding the resulting polynomial, we find
$$f(x-1) = 0.346749\, +0.520124 x+0.130031 x^2+0.00309598 x^3$$
So the probability of no matches among the Top Six teams is $0.346749$, the probability of exactly one match is $0.520124$, the probability of two matches is $0.130031$, and the probability of three matches is  $0.00309598$.
It is pleasant, but not essential, to have a computer algebra system to perform the polynomial expansion.
