Three girls $$G_1, G_2, G_3,$$ each read four stories $$ S_1, S_2, S_3, S_4 $$ and discuss which ones they like. No story is liked by all the three. For each of the three pairs of the girls, there is at least one story which is liked by the pair and not liked by the third. Let $$n$$ be the number of ways in which this is possible. Find the sum of the squares of the digits of $$n$$.

Label the three girls $$G_1,G_2,G_3$$ and the four stories $$S_1,S_2,S_3,S_4$$.
For every story record, in the same order $$G_1,G_2,G_3$$, who likes it (1 = likes, 0 = does not).
Thus every story is represented by a 3-digit 0-1 string.

The string $$111$$ (liked by all three) is forbidden by the first condition, whereas any of the other seven strings are allowed:

Pairs only (each must appear at least once, second condition):
$$A = 110,\; B = 101,\; C = 011$$

Singletons: $$100,\;010,\;001$$

Liked by none: $$000$$

Because the stories themselves are different, choosing strings for $$S_1,S_2,S_3,S_4$$ is an ordered assignment.
We have to count those assignments in which every one of the three pair-patterns $$A,B,C$$ appears at least once.

Let $$U$$ be the set of all possible assignments.
Since each story has 7 available patterns, $$|U| = 7^4 = 2401$$.

Define the following “bad” events:

$$E_A:$$ no story is of type $$A = 110$$
$$E_B:$$ no story is of type $$B = 101$$
$$E_C:$$ no story is of type $$C = 011$$

The required count $$n$$ equals the number of assignments in $$U$$ that avoid all three bad events. Use the Principle of Inclusion-Exclusion (PIE).

• If $$A$$ is forbidden, each story may take any of the remaining 6 patterns, hence $$|E_A| = 6^4$$, and likewise $$|E_B| = 6^4,\; |E_C| = 6^4$$.

• If both $$A$$ and $$B$$ are forbidden, only 5 patterns are left, so $$|E_A \cap E_B| = 5^4$$. All three pairwise intersections have the same size.

• If $$A,B,C$$ are all forbidden, only the 4 patterns $$000,100,010,001$$ remain, giving $$|E_A \cap E_B \cap E_C| = 4^4$$.

Applying PIE,

$$ \begin{aligned} n &= 7^4 - 3\cdot 6^4 + 3\cdot 5^4 - 4^4 \\[4pt] &= 2401 - 3(1296) + 3(625) - 256 \\[4pt] &= 2401 - 3888 + 1875 - 256 \\[4pt] &= 132. \end{aligned} $$

The digits of $$n$$ are 1, 3 and 2. Their squared values are $$1^2 = 1,\; 3^2 = 9,\; 2^2 = 4$$. Hence the required sum is $$1 + 9 + 4 = 14$$.

Answer: 14