0

I came across a question yesterday about combinations, and I wanted to know what the correct answer was. The question states as follows:

There are 8 spaces that are alternately black and white. There is one king, one queen, two identical rooks, two identical bishops, and two identical knights. The king needs to be surrounded by the two rooks. Then, the two bishops can be put on any of the remaining spaces as long as one is on a black space and the other is on a white space. Finally, the queen and the knights can be placed anywhere in the remaining spaces in relation to the other pieces. How many possible arrangements are there considering these conditions?

mariepoppins
  • 1
  • What is the significance of placing the rooks? – Hawk Jan 14 '14 at 21:13
  • 1
    Do the rooks have to be next to the king? – TonyK Jan 14 '14 at 21:14
  • Yes, each rook needs to be on either side of the king – mariepoppins Jan 14 '14 at 21:17
  • What was the source of the question? I surmise that the original question was about the number of legal starting positions in Fischer Random Chess aka Chess 960, in which case the requirement for placint the king and rooks is that the king must be somewhere between the two rooks (not necessarily right next to them), as in the normal array, so that some kind of castling is possible. The name "Chess 960" probably contains a hint as to what the answer is. – bof Jan 14 '14 at 21:25
  • @bof You are probably correct, I may have misinterpreted the rook/king situation. I don't have the source of the question, as it was on a test and I only wrote the question from memory. – mariepoppins Jan 14 '14 at 21:44
  • Maybe the instructor who devised the test misinterpreted it. I just think that, somewhere in the prehistory of the question, it started out as a question about Fischer Random Chess. – bof Jan 14 '14 at 21:48
  • @Hawk The reason for placing the rooks on either side of (but not necessarily right next to!) the king is so that castling will be somewhat the same as in orthodox chess. – bof Jan 14 '14 at 22:06

2 Answers2

3

Assuming that the king and rooks do not have to be on consecutive squares, the question is about the number of legal arrays in Fischer Random Chess aka Chess960. The answer is (surprise!) $960$. Here is an explanation from the Wikipedia page:

Each bishop can take one of four positions, the queen one of six, and the two knights can assume five or four possible positions respectively. This leaves three open squares which the king and rooks must occupy according to setup stipulations, without choice. This means there are 4×4×6×5×4 = 1920 possible starting positions if the two knights were different in some way. However, the two knights are indistinguishable during play (if swapped, there would be no difference), so the number of distinguishable possible positions is half of 1920, or 1920/2 = 960. (Half of the 960 are left-right mirror images of the other half, however Chess960 castling rules preserve left-right asymmetry in play.)

Of these $960$ positions, just $108$ satisfy the further condition that the rooks be right next to the king. Namely, there are $6$ ways to place the RKR combo. Any placement of the RKR combo leaves three squares of one color and two squares of the other color, so there are $3\times2$ ways to place the bishops. The queen then has a choice of $3$ squares, and the knights take what's left; $6\times3\times2\times3=108$.

The requirement that the king be placed between the rooks is natural, because it permits a kind of generalized castling with an effect somewhat similar to castling in orthodox chess. Requiring the rooks to be placed right next to the king has no apparent chessic motivation.

bof
  • 78,265
0

Hint: how many ways are there to select the leftmost square of the RKR series (I am taking surrounded by as one rook is immediately next to the king on each side)? Now place the bishops-how many ways to do that? Now place the queen-how many ways? The knights are now fixed.

Ross Millikan
  • 374,822
  • So since the question mentioned that the bishops, rooks, and knights are identical, then does that mean if I switched the placement of the rooks, but the king was in the same spot, it would still only count as one arrangement? – mariepoppins Jan 14 '14 at 21:22
  • That is correct. Similarly for the other pairs of pieces. – Ross Millikan Jan 14 '14 at 21:24
  • What I have is... Rook and king arrangements: 6 Bishops arrangements: 6 Queen arrangements: 3 Knights: 0 (Because they depend on the queen) Would the answer be 6 x 6 x 3 = 108? – mariepoppins Jan 14 '14 at 21:28
  • Yes, that is correct under my reading of the rook/king restriction – Ross Millikan Jan 14 '14 at 21:34
  • @mariepoppins Yes, $108$ if the king and rooks have to be on three consecutive squares, but I'm sure that's a misinterpretation of the original question. Under the rules of Fischer Random Chess aka Chess 960 there are $960$ positions: $4\times4=16$ ways to place the two Bs, then $\binom63=20$ ways to playe the RKR (not necessarily on adjacent squares), then $3$ squares for the Q, the two remaining squares for the knights, so $16\times20\times3=960$ ways. – bof Jan 14 '14 at 21:40
  • @RossMillikan Thank you! – mariepoppins Jan 14 '14 at 21:45
  • @bof Yes I probably have misinterpreted the rook and king situation. Thank you! Considering the condition of the rooks and the kings that you mentioned, why are there 16 ways to place the two bishops? – mariepoppins Jan 14 '14 at 21:47
  • There are $16$ ways to place the two bishops because they must go on different color squares, so pick one of the $4$ black squares and one of the $4$ white squares. (The two bishops are identical.) – bof Jan 14 '14 at 21:51