Ireland

First Solution: Represent each "score sheet" by a 8-digit ternary string with digits from $\{0,1,2\}$. Without loss of generality we may assume that Ann's score sheet reads 00001111, and that Barbara's score sheet reads 00002222. The total number of possible score sheets is $3^8$. We will count the number of score sheets which could not have been allocated to David: call this number $N$. Denote by $S_A$ the set of 8-digit ternary strings which differ from Ann's score in at most 3 places, and by $S_B$ the set of 8-digit ternary strings which differ from Barbara's score in at most 3 places. Then $$|S_A|=|S_B|=\left(\genfrac{}{}{0ex}{}{8}{0}\right)2^0+\left(\genfrac{}{}{0ex}{}{8}{1}\right)2^1+\left(\genfrac{}{}{0ex}{}{8}{2}\right)2^2+\left(\genfrac{}{}{0ex}{}{8}{3}\right)2^3=577.$$ Next we count $|S_A\cap S_B|$. For $s\in |S_A\cap S_B|$ there are two possible cases: The first 4 digits of $s$ are all zero, and the second 4 digits contain at least one 1 and at least one 2. There are $3^4-(2^4+2^4-1)=50$ of these, since $2^4$ contain no 1s, $2^4$ contain no 2s, and 1 contains no 1s and no 2s. Three of the first 4 digits of $s$ are zero, and the second 4 digits contains exactly two 1s and exactly two 2s. There are $4\cdot 2\cdot \left(\genfrac{}{}{0ex}{}{4}{2}\right)=48$ of these. Therefore $|S_A\cap S_B|=50+48=98$, and so $$N=|S_A\cup S_B|=|S_A|+|S_B|-|S_A\cap S_B|=2\cdot 577-98=1056,$$ leaving the number of possible score sheets for David as $3^8-N=5505$.

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Alternative solution.

Second Solution: Represent each score sheet by a 8-digit ternary string with digits from $\{0,1,2\}$. Without loss of generality we may assume that Ann's score sheet reads 00001111, and that Barbara's score sheet reads 00002222. Note that the number of differences between David's and Ann's scores in the first 4 digits is equal to the number of differences between David's and Barbara's scores in the first 4 digits. Next, note that the number of 4-digit ternary strings containing $r$ nonzero elements is given by $a(r)=\left(\genfrac{}{}{0ex}{}{4}{r}\right)2^r$. Also, denote by $b(r)$ the number of 4-digit ternary strings which differ from 1111 and 2222 in at least $r$ digits, for $r=0,1,2,3,4$. Then $b(0)=3^4$ (all strings are allowed) $b(1)=3^4-2$ (since all strings except 1111 and 2222 are allowed) $b(2)=3^4-2^4-2$ (since disallowed strings are those which differ from 1111 and 2222 in at most one place) $b(3)=1+\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot 2+\left(\genfrac{}{}{0ex}{}{4}{2}\right)\cdot 2=21$ (partitioning the count on the number of zeros in the string) * $b(4)=1$ (only the string 0000 is allowed). The total number of score sheets for David is then (partitioning the count according to the number of nonzero elements in the first 4 digits) $$\begin{array}{rl}N& =a(4)b(0)+a(3)b(1)+a(2)b(2)+a(1)b(3)+a(0)b(4)\\ & =\left(\genfrac{}{}{0ex}{}{4}{4}\right)2^4\cdot 3^4+\left(\genfrac{}{}{0ex}{}{4}{3}\right)2^3\cdot (3^4-2)+\left(\genfrac{}{}{0ex}{}{4}{2}\right)2^2\cdot (3^4-2^4-2)\\ & +\left(\genfrac{}{}{0ex}{}{4}{1}\right)2^1\cdot 21+\left(\genfrac{}{}{0ex}{}{4}{0}\right)2^0\cdot 1\\ & =3^4\left(\left(\genfrac{}{}{0ex}{}{4}{4}\right)2^4+\left(\genfrac{}{}{0ex}{}{4}{3}\right)2^3+\left(\genfrac{}{}{0ex}{}{4}{2}\right)2^2\right)-\left(\genfrac{}{}{0ex}{}{4}{3}\right)2^4-\left(\genfrac{}{}{0ex}{}{4}{2}\right)2^6-\left(\genfrac{}{}{0ex}{}{4}{2}\right)2^3+8\cdot 21+1\\ & =3^4(3^4-9)-2^6-3\cdot 2^7-3\cdot 2^4+169\\ & =3^8-(3^6+2^6+2\cdot 2^3\cdot 3^3)+169\\ & =3^8-(3^3+2^3{)}^2+13^2\\ & =5505,\end{array}$$ i.e., the number of score sheets possible for David is 5505.