سوال ۱۱۶ کنکور ارشد علوم کامپیوتر ۱۳۹۸

We must place $7$ identical, mutually non-attacking rooks on an $8\times 8$ board. Non-attacking means no two share a row or a column, so the $7$ rooks occupy $7$ distinct rows and $7$ distinct columns.

Step 1 — choose the rows. Pick which $7$ of the $8$ rows are used: $\binom{8}{7} = 8.$

Step 2 — choose the columns. Pick which $7$ of the $8$ columns are used: $\binom{8}{7} = 8.$

Step 3 — match rows to columns. The $7$ chosen rows must be matched bijectively to the $7$ chosen columns (one rook per chosen row, in a distinct chosen column). That is a permutation of $7$ objects: $7! .$

Since the rooks are identical, no extra factor for ordering the pieces is needed. By the multiplication principle: $\binom{8}{7}\binom{8}{7}\,7! = 8 \cdot 8 \cdot 7! = 64 \cdot 7!.$

Now rewrite this to match an option. Since $8! = 8\cdot 7!$, $8 \cdot 8! = 8 \cdot (8\cdot 7!) = 64 \cdot 7!,$ so the count equals $8 \times 8!$ — option 3.

Numerically, $8\times 8! = 8\times 40320 = 322{,}560$, which agrees with $64\times 5040 = 322{,}560$.

Exam trick: For $k$ non-attacking rooks on an $n\times n$ board the count is $\binom{n}{k}\binom{n}{k}\,k!$ (choose rows, choose columns, then a bijection). Here $k=7,\,n=8$ gives $8\cdot 8\cdot 7! = 8\cdot 8!$. The tempting wrong answer $8!$ forgets that one column is also left empty (it counts only the row choice × permutation), and $7\times 7!$ forgets one of the two free choices entirely.