# Math Help - Combinations brain twister

1. ## Combinations brain twister

This problem drives me crazy. The legend says is a 5th grade problem. I doubt.

- 10 marbles (0 to 9)
- 3 bowls.
- Each bowl can hold 3 non-consecutive marbles.
How many different combinations?

I have managed to solve it using a computer and brute force. I have solved an easier variant (5/2/2) without the help of a computer.

Thank you.

P.S. I have finished the school a long time ago, be gentle.

2. We need some clarifications.
Are the bowls identical or are they distinct?
You are putting nine balls into three bowls, three balls per bowl. Is that correct?

3. I hope this example will make things clearer:

Valid combination: 023 145 769. The order of bowls does not count, i.e. 145 769 023 is the same. Also the order of marbles in a bowl does not count, i.e. 451 976 203 is the same.

Invalid combination: 201 845 367. The marbles in first bowl are successive.

Invalid combination: 023 045 769. 0 is repeated.

4. Let's be clear: $\boxed{023}\;,\;\boxed{486}\;,\;\boxed{975}$ would be valid.
Note that 23 in the bowl.

5. Yes, 023 is valid. 123 or 231 is not valid.

6. ## Solution?

The count of successive-marbles combinations is:
S( R, M ) = R – M + 1
where
R is the total number of marbles;
M is the number of marbles a bowl is holding.
In our case S( 10, 3 ) = 8.

Let B be the number of bowls.

Method 1. Remove the successive-marbles combinations from start.
1. The count of M marbles combinations is: U = C( R, M ) (120, in our case)
2. The count of M marbles combinations, except for the successive-marbles combinations is V = C( R, M ) – S( R, M ) (112, in our case)
3. The count of B bowls combinations (bowls containing only non-successive marbles) is W = C( V, B ) (227920, in our case).
4. 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.
Method 2. Remove the identical marble combinations right from the start.
1. 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).
2. 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.

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.