The count of successive-marbles combinations is:
S( R, M ) = R – M + 1where
R is the total number of marbles;In our case S( 10, 3 ) = 8.
M is the number of marbles a bowl is holding.
Let B be the number of bowls.
Method 1. Remove the successive-marbles combinations from start.
Method 2. Remove the identical marble combinations right from the start.
- The count of M marbles combinations is: U = C( R, M ) (120, in our case)
- The count of M marbles combinations, except for the successive-marbles combinations is V = C( R, M ) – S( R, M ) (112, in our case)
- The count of B bowls combinations (bowls containing only non-successive marbles) is W = C( V, B ) (227920, in our case).
- From W set we now have to remove the combinations that contain the same marble more than once. That is a dead end; for me, at least.
- The count of non-identical-marble combinations is the first bowl is C( R, M ); the number of combinations of the remaining R – M marbles is C( R – M, M ); and so on until we fill all B bowls. The total number of combinations is C( R, M ) * C( R – M, M ) * … Since the bowls may change places we have to divide it by B!. That is: Q = C( R, M ) * C( R – M, M ) * … / B! (2800 in our case).
- Now remove the successive combinations. Someone I know managed to compute this number for our case (504), but did not manage to come up (yet) with a general equation.
Comments are welcome.
a. First put 012 in the first bowl. Following the same reasoning at 1. we find 70 combinations.
b. Same for 123: 70 combinations; 012 cannot be among those combinations.
c. Same for 234: 70 combinations; 012 and 234 cannot be among those combinations.
d. 345: 66 combinations; the total number of 345 combinations is 70 but 012 occurs 4 times, so we have to subtract 4.
e. Follow the same reasoning at d. until we reach the 789 combination.
f. 70 + 70 + 70 + 66 + 62 + 58 + 55 + 53 = 504
g. ANSWER = 2800 – 504 = 2296, that is the same answer I got using brute force.