Question:

Twenty five coloured beads are to be arranged in a grid comprising of five rows and five columns. Each cell in the grid must contain exactly one bead. Each bead is coloured either Red, Blue or Green.
While arranging the beads along any of the five rows or along any of the five columns, the rules given below are to be followed:
(1) Two adjacent beads along the same row or column are always of different colours.
(2) There is at least one Green bead between any two Blue beads along the same row or column.
(3) There is at least one Blue and at least one Green bead between any two Red beads along the same row or column.
Every unique, complete arrangement of twenty five beads is called a configuration.
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?
[This Question was asked as TITA]

Updated On: Jul 24, 2025
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The Correct Option is A

Approach Solution - 1

To solve this problem of arranging colored beads following the given rules and maximizing the number of Red beads, we need to break down the problem systematically: 
(1) According to Rule 1, no two adjacent beads can have the same color. This implies a checkerboard pattern limitation across rows and columns.
(2) Rule 2 requires at least one Green bead between two Blue beads, ensuring the alternation constraint remains.
(3) Rule 3 necessitates both at least one Blue and one Green bead between two Red beads, increasing color diversity.
Given placements: Red at 'second row, third column' and 'third row, second column'. We explore one possible optimal placement configuration maximizing Red beads within constraints:

GBRBG
BRGRB
RGGGB
BRBRG
GBGBR

In this configuration:

  • Each of the two initially placed Red beads adheres to rules. New Red beads are placed at positions carefully, aligned within constraints, namely (2,4), (4,2), (4,4), and (5,5) maximizing the number of Reds.
  • The grid maintains compliance with all rules, particularly every Red's required alternation with Blue and Green.
  • Counting reveals a total of 8 Red beads, with 2 given, 6 additional Red beads placed legally and optimally.

Therefore, the maximum additional Red beads possible is 6.

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Approach Solution -2

6 more beads can be placed as shown below:
Arrangement
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