سوال ۱۱۳ کنکور ارشد علوم کامپیوتر ۱۴۰۳

We count strings of length $10$ over the alphabet $\{1,2,3\}$. Define two events:

• $A$: the five leftmost digits contain no $1$ (each left digit $\in \{2,3\}$).
• $B$: the five rightmost digits contain no $2$ (each right digit $\in \{1,3\}$).

We want $|A \cup B|$ and use inclusion-exclusion: $|A \cup B| = |A| + |B| - |A \cap B|.$

Count $|A|$. The 5 left positions each have $2$ choices ($\{2,3\}$) and the 5 right positions are unrestricted ($3$ choices each): $|A| = 2^{5} \cdot 3^{5}.$

Count $|B|$. Symmetrically, the 5 right positions each have $2$ choices ($\{1,3\}$) and the 5 left positions are free: $|B| = 3^{5} \cdot 2^{5}.$

Count $|A \cap B|$. Now the left 5 are restricted to $\{2,3\}$ and the right 5 to $\{1,3\}$. These constraints touch disjoint positions, so $|A \cap B| = 2^{5} \cdot 2^{5} = 2^{10}.$

Combine. $|A \cup B| = 2\cdot 2^{5} 3^{5} - 2^{10} = 2^{6}\cdot 3^{5} - 2^{10}.$

Since $2^{10} = 2^{6}\cdot 2^{4} = 2^{6}\cdot 16$ and $3^{5} = 243$, $|A \cup B| = 2^{6}\bigl(243 - 16\bigr) = 2^{6}\cdot 227.$

Numerically $64\cdot 243 - 1024 = 15552 - 1024 = 14528 = 64\cdot 227$. This is option 3.

Pitfall: Do not just add $|A|+|B|$ (which gives $2^{6}\cdot 3^{5}$, the near-miss option 4 idea) — the overlap $A\cap B$ is double-counted. Subtracting $|A\cap B| = 2^{10}$ turns $243$ into $243-16 = 227$.