In the Gadsden Diving Championships there is a panel of three judges. There are 12 scorecards with the numbers 1-12 on them, which are distributed so that each judge has four. After the diver has performed their routine, the judges each hold up a score card, and the diver's score is the sum of the three scorecards added together.
So, my question is:
What are all the different ways that the cards can be distributed among the judges, if all player scores from 6-29, except 10, were possible? Also, explain why ithere's no other way of distributing the scorecards. (other than swapping sets of 4 scorecards between judges)
Working Out:
I've already worked out some combinations, such as
Are there any others? Thanks!
To make $6$ possible, $1$, $2$ and $3$ need to be in different hands. (Let's label the judges $J_i$ according to which of these they hold, as you did in your image.)
The only combination for $7$ is $1+2+4$; so $J_3$ must have $4$.
The only combinations for $8$ are $1+2+5$ and $1+3+4$; the second one is no longer possible, so $1+2+5$ must be possible, so $J_3$ must have $5$. The only combinations for $9$ are $1+2+6$, $1+3+5$ and $2+3+4$. The last two are no longer possible, so $J_3$ must have $6$. Thus we've determined $J_3$'s hand.
Now for $11$ to be possible, we need a combination from $J_1$ and $J_2$ that sums to at most $8$ (other than $1+2=3$). Your first row fails in this regard; it doesn't generate $11$. This is only possible if we give the $7$ to $J_2$. This also makes $12$ to $14$ possible. So far, we have:
$$ J_1: 1\;,\\ J_2:2,7\;,\\ J_3:3,4,5,6\;. $$
$J_3$'s highest card is $6$, so for $29$ to be possible, $J_1$ and $J_2$ must each have one of $11$ and $12$. Then $26$ to $28$ are also possible.
That leaves only one card to give to $J_2$, and it can be $8$, $9$ or $10$. Together with a $6$, $5$ or $4$ from $J_3$ and the $1$ from $J_1$, that yields $15$.
The $1$ from $J_1$ with the $11$ or $12$ from $J_2$ adds to $12$ or $13$, and with $J_3$'s cards that gets us $16$ through $18$.
The $7$ from $J_2$ with the $8$ or $9$ from $J_1$ adds to $15$ or $16$, and with $J_3$'s cards we get $19$ through $21$.
The $7$ from $J_2$ with the $11$ or $12$ from $J_1$ adds to $18$ or $19$, and with $J_3$'s cards that yields $22$ through $24$.
The $9$ from either $J_1$ or $J_2$ with the $11$ or $12$ from the other of the two adds to $20$ or $21$, and with a $5$ or $4$ from $J_3$ that yields $25$, and we're done.
Thus, there are exactly $2\cdot3=6$ admissible combinations, and you get them by starting from the combination displayed above, giving $J_1$ and $J_2$ one each of $11$ and $12$, then independently choosing one of $8$, $9$ and $10$ to give to $J_2$ and then giving the other two of those to $J_1$.