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Anyway, these two simulated race are less likely to feel more and write it today preliminaries nice dl
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Anyway, the topic to remind you that you excatalan Cattleya number of m = 0 is the Cattleya you find
Then consider m is any number of methods are clearly proof we need to know the number of Cattleya fact, yesterday when I wanted to explore more ways to solve this problem it is necessary
First, there is proof that we might fold line leads from the fold line is another proof idea
Cattleya time corresponding to the number of problems models are performing operations in the K-2 operations are executed at least 1 K times so we use the x-axis of the current total number of operations required 2n times
Then the operation as upward 1 45 2 2-step operation as a downward root root of 2 step 45 then down then we can find a route to (2n, 0) of the fold line
So the number of legal program we can find that is the broken line does not cross the x-axis at any time we consider this time minus the number of illegal use of several programs all programs
At this time, the legitimate sequence corresponds exactly to the number of brackets we find legitimate program paths then all programs corresponding to the number of $ \ binom {2n} {n } $
Then the time to consider how to do everything illegal number of programs that we remember what we seek from (0,0) starting up or to the right to reach the (n, n)
Not the number of programs across the y = x is how we verify it we draw a map to explore what I really do not want to draw cut combinatorics on yxc fairy b explain station
Now there is not a legitimate path to any we find for the first time crossed the line to reach another part of a line that is part of the green on the map
From this point to this point we are back on the straight path all folded in the past
At this time, the end point then becomes (n-1 n + 1) then all invalid paths corresponding to all data on the number of all paths that arrive from the program (0,0) (n-1, n + 1) we clearly understood
At this point we return to the above method we consider the fold line 45 each time the name of the method is the first x-axis crosses the straight line y = -1 arrival time between the path K K after 2n + 1 ~ y on all = -1 symmetrically over
However, this is to the reference line y = x and the x-axis becomes green line
Then the end becomes (2n, -2) where operation 1 less than 2 then the operator operating an operation has 2 n-1 n + 1 has then consider this case is the number of selected programs in the 2n n-1 thereof $ \ Binom {2n} {n-1} $
So the number of illegal program number is the number of Cattleya program at least once a mismatch at this time we have proved out the need to find ways to put this folding baseline understanding of it.
At this time, $ \ binom {2n} {n} - \ {n-1} $ is the number of programs from the fact this time we might think binom {2n} is what we subtract the number corresponding to the at least one program that does not match
If we consider the exactly m times the number of illegal program then we may be converted into at least m number of programs that do not match - at least m + 1 number of times corresponding to the program actually does not match to the line chart
In this case we put into the reference line y = -m This time we find out the number of the above-described embodiment according to the number of time we find that at least m out illegal program
At this time, CAT (n-, m ) -cat (n-,. 1 + m) In this case the answer is the CAT (n-, m) = $ \ n-Binom {} {} 2N - \ Binom 2N} {} {$. 1-nm