There are two approaches, and I don't know if either is amenable to a pencil and paper methodology.
#0:
Let the pots be labeled a, b, c and d. Whatever order each pot's balls are chosen, theres a specific order in whith the pots' order will be chosen--some permutation of aaabbbcccddd. Each valid permutation must have no two successive letters match. At any given stage there are three choices, unless one or more has been used three times already. The depletion of a given letter may occur at different points in the process, so it isn't as simple as 4*3^11, or even an easy variation on that. So an idea would be to, for each stage, multiply by, not 3, but rather 3 times the fraction of the preceding portions of the permutation that did not deplete any of the remaining letters, etc. But these are not independent.
Of course a computer could examine each permutation to determine if it is a valid one. Perhaps this would not be disqualified on the prohibition of simulations, as this is only part of the overall solution; that would depend on what one considers to be a simulation. A simulation would use either a Monte Carlo technique or an exhaustive count of all possibilities. The exhaustive count we consider here is only of the permutations of aaabbbcccddd (and in later parts, such strings as aaaabbbcccddd), not of the entire set of possibilities.
For each ordering of the pots, there are 6^4 overall ways of choosing all 12 balls, as there are 6 orders for choice of balls within a pot for the three occurrences of each of the four pots.
So the final answer would be the number of valid permutations of aaabbbcccddd multiplied by 6^4.
#1:
Well, this time there are 13 selection from among, say aaaabbbcccddd. Once the valid permutations of this string are counted, we first must multiply by 4, as any of the four pots may be chosen for a repetition.
Then, within the choice orders for a given pot, there are the usual 6 (i.e., 3!) for primary choice. In each of the pots not chosen for a replacement this number stands. This time there are only three such non-replacement pots, accounting for a multiplication by 6^3.
Now we factor in the possibilities for the pot in which replacement occurs. Again there are 6 orders of choice, and now if the first one chosen is replaced it may be chosen again in any of the three positions thereafter; the second one, only two possible positions; the last one, only 1. In summary, for balls 1, 2 and 3:
1123, 1213, 1231, 1223, 1232, 1233 are the possible orderings, for the initial ordering 123. As there are 6 initial orderings the total possibility is 6*6=36.
To summarize: the valid permutations of aaaabbbcccddd must be multiplied by 4 * 6^3 * 36.
Parts #2 and #3 are so complicated they wouldn't be fun any more, at least for me.
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Posted by Charlie
on 2020-02-12 19:01:48 |
outline of a solution methodology
