Small question: In the paper by Bayer and Diaconis "TRAILING THE DOVETAIL SHUFFLE TO ITS LAIR", the authors claim that the number of ways to riffle shuffle a certain ordering of cards $\pi$ from a completely ordered deck of $n\geq 2$ cards is $$\binom{n+a-r}{n}.$$ Thereby $r$ is the number of maximal rising sequences in $\pi$ and $a$ is the number of packs that the initial deck has been split into.
However, I am struggling with the result for $a=3$ (packs can be empty!) and $r=2$. Let's assume $\pi=1\, 2\, 3\, 4\, 7\, 5\, 8\, 6\, 9\, 10$. The two rising sequences are $1,2,3,4,5,6$ and $7,8,9,10$. A riffle shuffle is order presevering, so the deck must have been split between six and seven (that gives two packs).
The formula above gives me $$\binom{10+3-2}{10}=\binom{11}{10}=11.$$
However, i could either split the first pack in 7 places (remember, the empty set is a possibility) or the second in 5 places. That is a total of $$\binom{7}{1}\binom{5}{0}+\binom{7}{0}\binom{5}{1}=12\neq 11.$$
Do I have to assume, that the first pack has at least one card?
However, if i assume $a=4$ instead, i get $\binom{10+4-2}{10}=66$ and also $$\binom{7}{2}\binom{5}{0}+\binom{7}{1}\binom{5}{1}+\binom{7}{0}\binom{5}{2}=21\cdot 1+7\cdot 5+1\cdot10=66.$$
I can't seem to bring that together...