Probability of being dealt a bridge hand with... Hey Guys I'd like to know if this question can be tackled this way.
What is the probability of being dealt a bridge hand with exactly 4 honour cards and exactly 4 cards from the 5 through 10? 
I wrote that having in mind that the 10 is not included as it wasn't specified.
$$\frac{\binom{20}{4}\binom{20}{4}\binom{12}{5}}{\binom{52}{13}}$$
I wasn't really sure about choosing between the two fours. For example knowing which one is what. For that matter I wanted to go with the multinomial way.
$$\frac{\binom{40}{4,4}\binom{12}{5}}{\binom{52}{13}}$$
Any clarification would be appreciated.  

We split the bridge cards into $4$ mutually exclusive sets:


*

*$A$ is the set of $10$s; it consists of $4$ cards.

*$B$ is the set of honour cards which are not $10$s; it consists of $16$ cards.

*$C$ is the set of cards with number $5$ through $9$; it consists of $20$ cards.

*$D$ is the set of cards which are not in any of the sets above; it consists of $52-4-16-20=12$ cards.


Consider that a honour card is a card in the set $A$ or $B$, and a card with number $5$ through $10$ is a card in the set $A$ or $C$. That means we pick exactly $4$ cards in $A\cup B$ and exactly $4$ cards in $A\cup C$. That implies the number of cards we pick from $B$ and $C$ are the same.
Now we split cases on the number of members of $A$ picked:


*

*Case 1: $0$ in $A$, so $4$ in $B$, $4$ in $C$ and $5$ in $D$. There are $\binom{4}{0}\binom{16}{4}\binom{20}{4}\binom{12}{5}$ possibilities.

*Case 2: $1$ in $A$, so $3$ in $B$, $3$ in $C$ and $6$ in $D$. There are $\binom{4}{1}\binom{16}{3}\binom{20}{3}\binom{12}{6}$ possibilities.

*Case 3: $2$ in $A$, so $2$ in $B$, $2$ in $C$ and $7$ in $D$. There are $\binom{4}{2}\binom{16}{2}\binom{20}{2}\binom{12}{7}$ possibilities.

*Case 4: $3$ in $A$, so $1$ in $B$, $1$ in $C$ and $8$ in $D$. There are $\binom{4}{3}\binom{16}{1}\binom{20}{1}\binom{12}{8}$ possibilities.

*Case 5: $4$ in $A$, so $0$ in $B$, $0$ in $C$ and $9$ in $D$. There are $\binom{4}{4}\binom{16}{0}\binom{20}{0}\binom{12}{9}$ possibilities.


Now we add those numbers and divide by $\binom{52}{13}$.
