In Q1, Q, and Q4 above, ought not we multiply the final answers by 2, as there are 2 colors viz. black and white (and for each of these 2 the arrangements should be counted as discrete)?
Please correct me if I am wrong.

In the last question, shouldn’t we divide by (3!/2!) = 3 mathematically speaking. Also, logically speaking, the square for the white piece can be chosen in 3 ways, so we divide by 3.

If you had just chosen 3 squares initially, then you would multiply by 3 at the end for the position of the white. That would give you the same answer by a somewhat longer route as it would have been (64 * 49 * 36) / (3*2*1) and then *3.

In Q1, Q, and Q4 above, ought not we multiply the final answers by 2, as there are 2 colors viz. black and white (and for each of these 2 the arrangements should be counted as discrete)?

Please correct me if I am wrong.

You are not asked to choose the colour. You are given two pawns of the same colour and told “place them”.

regards

J

Okay, Thanks !

In the last question, shouldn’t we divide by (3!/2!) = 3 mathematically speaking. Also, logically speaking, the square for the white piece can be chosen in 3 ways, so we divide by 3.

If you had just chosen 3 squares initially, then you would multiply by 3 at the end for the position of the white. That would give you the same answer by a somewhat longer route as it would have been (64 * 49 * 36) / (3*2*1) and then *3.

regards

J