(1) When there are 179 red balls, (there are 179 subscript 36 1 release modes, 361× 360× …×183//kloc-0. )
When you put the blue ball and the gray ball together again,
Regardless of the numbering order, there are 178 playback modes. (The blue ball is 1 ~ 178, and the rest are gray balls. Note: There must be at least 1 blue balls, but not 0, otherwise there will be spaces. )
Consideration number: red ball has used 179, leaving 182. Put the blue ball first. After the blue ball is put away, the gray ball can only be put in the rest of the grid (the way to put the gray ball is unique).
Blue ball 1 gray ball 18 1, with c superscript 1 subscript 182.
There are two gray balls (180) in the blue ball, and there are two ways to put them on C with superscript 2: 182.
……
There are four blue balls, gray balls 178, and there are 178 subscripts 182 ways to play.
C superscript 1 subscript 182+C superscript 2 subscript 182+C superscript 3 subscript 182+...+C superscript 178 subscript 182.
= (2 182-C subscript 0 182-C subscript 182-C subscript 180 subscript 182-C subscript 1 subscript 60.
=(2^ 182- 182× 18 1× 180/3! - 182× 18 1/2! - 182-2)
So this situation * * * has a combination number: (2182-182 ×181×180/3! - 182× 18 1/2! -182-2)×C superscript 179 subscript 36 1
(2) When putting 180 red ball,
When you put the blue ball and the gray ball together again,
Regardless of the serial number, there are 179 playback modes. (put 0 ~ 178 for the blue ball, and put the rest in the gray ball. )
Considering the number, the combination number = (= (2 18 1-C+0-c superscript 179 subscript18 1-C superscript18/.
=(2^ 18 1- 18 1× 180/2! -18 1- 1)×C superscript 180 subscript 36 1
(3) When there are 18 1 red balls,
Regardless of the serial number, there are 179 playback modes as above. (put 0 ~ 178 for the blue ball, and put the rest in the gray ball. )
Considering the number, the number of combinations = (2180-1) × c superscript 18 1 subscript 36 1.
(4) When there are 182 red balls,
Combination number = (2 179- 1) × C superscript 182 subscript 36 1.
(5) When there are 183 red balls,
Combination number = = 2178× c c superscript 183 subscript 36 1
(6) When there are 184 red balls,
Combination number = = 2177× c c superscript 184 subscript 36 1
(7) When there are 185 red balls,
Combination number = = 2176× c c superscript 185 subscript 36 1
……
(183) When there are 36 1 red balls, there are 1 combinations. (0 blue balls and 0 gray balls)
Combination number = 2 0× c superscript 36 1 subscript 36 1= 1.
Add up the above to get the number of all possible combinations and permutations.
The knowledge of permutation and combination has been far away from me for a long time. I wonder if there is a formula that can simplify these formulas. Anyway, I won't, Khan ...)