Two Red beads have been placed in ‘second row, third column’ and ‘third row, second column’. How many more Red beads can be placed so as to maximise the number of Red beads used in the configuration?

The correct answer is 6.

This problem is based on a classic bead-placement puzzle on an 8 × 8 grid. The rule governing the configuration is that no two Red beads can share the same row or the same column. This ensures maximum spread and a valid configuration. The goal is to place as many Red beads as possible while respecting this rule.

Step 1 – Understand the Initial Placement

Two Red beads are already placed at:

  ��� Second row, Third column → Position (Row 2, Col 3)
  • Third row, Second column → Position (Row 3, Col 2)

Step 2 – Identify Blocked Rows and Columns

Since a Red bead occupies Row 2 and Row 3, no additional bead can be placed in those rows.
Similarly, since a Red bead occupies Column 2 and Column 3, no additional bead can be placed in those columns.

Blocked rows: Row 2, Row 3
Blocked columns: Col 2, Col 3

Step 3 – Identify Available Rows and Columns

Free rows (out of 8 total): Row 1, Row 4, Row 5, Row 6, Row 7, Row 8 → 6 rows
Free columns (out of 8 total): Col 1, Col 4, Col 5, Col 6, Col 7, Col 8 → 6 columns

Step 4 – Calculate the Maximum Additional Beads

We can place exactly one Red bead per free row, in a free column. Since there are 6 free rows and 6 free columns, we can place one bead in each of the 6 remaining rows (each in a distinct free column), satisfying the no-same-row, no-same-column rule.

Number of additional Red beads = 6

Step 5 – Verify the Total

Total Red beads in final configuration:

$2\left(\text{already placed}\right)+6\left(\text{newly placed}\right)=8\text{Red beads}$

This is the maximum possible for an 8 × 8 grid under the given constraint (one per row, one per column), similar in logic to placing 8 non-attacking rooks on a chessboard.

Conclusion: With the two Red beads already at (Row 2, Col 3) and (Row 3, Col 2), exactly 6 more Red beads can be added to the grid — one each in Rows 1, 4, 5, 6, 7, and 8 — to achieve the maximum configuration of 8 Red beads total.