$9$-card-long (or longer) straight flush in a $13$-card hand The other day at an online bridge site I got dealt a $13$-card hand containing $3,4,5,6,7,8,9,10,J$ of spades, i.e. a $9$-card-long straight-flush (s-f).  This is so rare that I decided to count how many $13$-card hands contain a s-f of length $9$ or longer.
For the purpose of this question:

*

*A s-f is a consecutive sequence in the same suit, and aces always count high (since I was playing bridge!) so $A,2,3,4,5$ of spades is not a $5$-card s-f for this purpose.


*Let us just count s-f in spades.  If we want to count all suits we can simply multiply by $4$ since a hand cannot contain two s-f's of different suits each of length $\ge 9$.
I was trying to use inclusion-exclusion but got very confused.  Finally I decided to count explicitly as follows.  My questions:
(1) Can someone verify my answer below?
(2) Can someone show how this can be counted via inclusion-exclusion, or explain why that's not a good idea?

My attempt:

*

*$13$-long s-f: No. of hands $=1$.


*exactly $12$-long s-f: There are two such ($2-K$ or $3-A$), and each must be completed with an extra card (to make a $13$-card hand) but the extra card must not complete a $13$-long s-f.  No. of hands $= 2 \times {39 \choose 1}$.


*exactly $11$-long s-f: Three such; the $2-Q$ hand must not include the $K$ (leaving $40$ valid cards to choose $2$ to fill out the hand), the $3-K$ hand must not include $2$ or $A$ (leaving $39$ valid cards), and the $4-A$ hand must not include the $3$.  No. of hands $= {40 \choose 2} + {39 \choose 2} + {40 \choose 2}$.


*exactly $10$-long s-f: Similar to above, no. of hands $= {41 \choose 3} + {40 \choose 3} + {40 \choose 3} + {41 \choose 3}$.


*exactly $9$-long s-f: Similar to above, no. of hands $= 2 \times {42 \choose 4} + 3 \times {41 \choose 4}$.
According to wolfram alpha total $= 571130$.
 A: Your count is correct, and here is the inclusion-exclusion approach, where the five properties correspond to the five straight flushes:
\begin{align}
&\binom{5}{1}\binom{52-9}{13-9} \\
&-\left[4\binom{52-10}{13-10}+3\binom{52-11}{13-11}+2\binom{52-12}{13-12}+1\binom{52-13}{13-13}\right] \\
&+\left[3\binom{52-11}{13-11}+4\binom{52-12}{13-12}+3\binom{52-13}{13-13}\right] \\
&-\left[2\binom{52-12}{13-12}+3\binom{52-13}{13-13}\right] \\
&+\binom{5}{5}\binom{52-13}{13-13} \\
&= 
617050
-48461
+2623
-83
+1 \\ 
&=571130
\end{align}
A: Inspired by the PIE-based answer of @RobPratt I found the following much simpler (computationally speaking) solution!
As usual all s-f's considered here are in spades.
Consider the value $5 \times {52-9 \choose 13-9}$.  This is the number of pairs of sets $(X,Y)$, where $|X| = 9$ and the contents form a $9$-long s-f and $|Y| = 4$ and the contents are arbitrary, as long as $X, Y$ are disjoint.  The first factor in the product is $5$ because there are $5$ possible $9$-long s-f's.
Now, every hand with an exactly $9$-long s-f can be represented as $(X,Y)$ in exactly one way.  Meanwhile, every hand with an exactly $10$-long s-f can be represented as $(X,Y)$ in exactly two ways, and every hand with an exactly $11$-long s-f can be represented as $(X,Y)$ in exactly three ways, etc.  Therefore:
$$5 \times {52 - 9 \choose 13 - 9} = N_9 + 2 N_{10} + 3 N_{11} + 4 N_{12} + 5 N_{13}$$
where $N_j$ is the no. of hands with an exactly $j$-long s-f.
By a completely analogous argument but this time considering $(W,Z)$ where $|W| = 10$ and the contents form a $10$-long s-f, and $|Z| = 3, W \cap Z = \emptyset$, we have
$$ 4 \times {52 - 10 \choose 13 - 10} = N_{10} + 2 N_{11} + 3 N_{12} + 4 N_{13}$$
The difference between the two gives the desired count, i.e. no. of hands with $9$-long (or longer) s-f:
$$N_9 + N_{10} + N_{11} + N_{12} + N_{13} = 5 \times {52 - 9 \choose 13 - 9} - 4 \times {52 - 10 \choose 13 - 10} = 571130 \,\,\,\,\,\,\square$$
