Note that there are two ways to generate valid paths in the first description when $$$n=m$$$, one keeps $$$x\ge y$$$ and the other keeps $$$x\le y$$$, so the answer is $$$\frac{2}{n+1}\binom{2n}{n}$$$.

If one buys a ticket with only a 20 euro banknote, Charlie should use a 10 euro banknote for change. Say Charlie has served $$$x$$$ customers with only 20 euro and $$$y$$$ with only 10 euro, the inequality $$$y\ge x-k$$$ must hold to keep the process run.

Therefore, the not-intersected line in this question is $$$y=x-k-1$$$ and $$$(0,0)$$$ reflects to $$$(k+1,-k-1)$$$, which gives the number of good permutations as:

And the answer to this problem is:

Suppose the printer does the operation 2 for $$$i$$$ times and does the operation 1 for the other $$$m-i$$$ times. Since the string only has $$$n$$$ characters, we must use $$$m-i-n$$$ operation 2 to clear extra characters. And now we have $$$k=i-(m-i-n)$$$ extra operations 2 which must be used on empty string to waste them.

Now we suppose the printer does $$$x$$$ operation 2 and $$$y$$$ operation 1 at some time, and we can find the same inequality $$$y\ge x-k$$$ also holds. In addition, the equality $$$y=x-k$$$ must hold at a certain time. In other words, the path must intersect with $$$y=x-k$$$ but cannot go through it.

To find the answer, we first find the number of non-intersect paths, which is $$$\binom{n}{i}-\binom{n}{i-k-1}$$$, and then find the number of non-go-through paths, which is $$$\binom{n}{i}-\binom{n}{i-k}$$$. Note that our paths are intersects but not goes through with the line, so the number of our paths is that of non-go-through ones minus non-intersect ones, or

The problem can be easily solved by enumerating $$$i$$$ and accumulating results.

Note that player 1 must have no less suit 1 cards (or trumps) than player 2, and Of each other suits, player 1 must have no more cards (or non-trumps) than player 2. Particularly, player 1 uses his small trumps to ruff extra non-trumps of player 2.

Say $$$m=2p$$$ and in a distribution, player 1 has $$$p-q$$$ cards of some suit other than 1, so player 2 has $$$p+q$$$ cards. We can list $$$1$$$ or $$$2$$$ according to the cards' ranks of two players. For example, if player 1 obtains $$${3,6}$$$ and player 2 obtains $$${1,2,4,5}$$$, the list is $$$221221$$$. To ensure each of player 1's cards finds an opponent's distinct card whose rank is below it, the list must satisfy that in each prefix of our list, the number of $$$2$$$ must be no less than $$$1$$$.

Therefore, we can just substitude $$$m=p-q$$$ and $$$n=p+q$$$ into the formula above and get the number of valid distributions of this suit:

For suit 1, the formula is the same, i. e., when player 1 has $$$p+q$$$ trump card and player 2 has $$$p-q$$$, there are also $$$a_q$$$ valid distributions.

Enumerate $$$q$$$ of suit 1, and the sum of $$$q$$$ of the other suits is equal to $$$q$$$ we enumerate. Therefore, the valid distributions of other suits is $$$a^{*(n-1)}[q]$$$, so the final answer is: