CAT 2020 [Slot 2] DILR set 1

As we need to use only two colours, in any row or column these two coloured beads will be placed alternately like

So we cannot place Red coloured beads at position 1 or two as between any two Red beads there must at least two beads (at least one green and at least one Blue). Hence, we can use only Green and Blue coloured beads.

We can have two possible configurations:

Configuration 1: Green bead is placed at top left corner

Configuration 2: Blue bead is placed at top left corner

Between Any two Red beads there must be at least two Beads. So any Row or column there can be maximum two red beads. If we place two red beads in each row then two columns will have three red bead which cannot be accepted.

The above configuration is not correct.

So in the third row we will place only one Red bead at the middle of the third row. Also we will adjust other rows so that between any two Red beads there are at least two beads in any column.

So maximum 9 Red beads are possible in any configuration. At remaining places Green and Blue coloured beads can be placed in such way that all the conditions given are satisfied. There are multiple configurations are possible. One of the configurations is given as below.

To minimise number of Blue beads we need to maximise number of Red and Green beads. From the previous question solution, Maximum no. Red beads can be 9. The row in which has two red beads, we will place two green and one Blue bead additionally.

The row with only one red bead we will place two green and two blue beads additionally. So overall there will be minimum 6 Blue beads.

We can place maximum 6 more beads as shown below.