IPMAT Indore 2023 · Quantitative Ability · Question 23

Step 1 — understand the constraint. There are 8 rooks and 8 rows. “No two in the same row” with 8 rooks on 8 rows forces exactly one rook per row. The same argument gives exactly one per column.

Step 2 — place them row by row.

Row 1: the rook may go in any of the 8 columns.

Row 2: any column except the one already used — 7 choices.

Row 3: 6 choices.

…

Row 8: 1 choice.

Step 3 — multiply.

8\times 7\times 6\times \cdots \times 1=8!

Step 4 — recognise the structure. Each placement corresponds to choosing which column each row uses, with no repeats — that is precisely a permutation of the 8 columns among the 8 rows.

8!=40320

Answer: Option 4 — 8!

Shortcut. This is a permutation in disguise. A valid placement is exactly a bijection from rows to columns, so the count is 8! — no case work, no board drawing.

The trap is option (b), 7!. It comes from thinking the first rook is somehow “fixed” (as it would be in a circular arrangement). There is nothing circular here: the board has a distinguishable first row and first column, so all 8 choices are genuinely different.

The general result: n non-attacking rooks on an n\times n board can be placed in n! ways. If instead you place only k<n rooks, the count is {\left( \genfrac{}{}{0pt}{}{n}{k}\right)}^{2}k! — worth knowing, since that variant appears often.