Question 1208189: How many possible are there to place two queens on an empty chessboard such that neither threatens the other?
Found 3 solutions by PPCHLOE, ikleyn, mccravyedwin:
Answer by PPCHLOE(2)   (Show Source):
You can put this solution on YOUR website! Hello here is the solution to your problem in steps.
Hope you find this helpful.
ogunjimio79@gmail.com
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Step 1: Understand the constraints
- Two queens cannot share the same row, column, or diagonal.
- A standard chessboard has 8 rows and 8 columns.
Step 2: Calculate the total possible placements for the first queen
- The first queen can be placed in any of the 64 squares (8 rows x 8 columns).
Step 3: Calculate the possible placements for the second queen
- After placing the first queen, the second queen cannot be placed in the same row, column, or diagonal.
- This means the second queen has 36 possible squares to be placed (64 - 28 squares that are under attack by the first queen).
Step 4: Account for overcounting
- Since the order of placing the queens doesn't matter, we need to divide the total count by 2 to avoid counting each configuration twice.
Step 5: Calculate the total possible ways
- Total possible ways = (Total possible placements for the first queen) x (Possible placements for the second queen) / 2
- Total possible ways = 64 x 36 / 2
- Total possible ways = 1152
The final answer is 1152.
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If you're satisfied with my solution and interested in further learning , I'm available for a one-on-one online session. I'd be happy to share my knowledge and provide guidance to you personally .
Answer by ikleyn(52787)  (Show Source):
Answer by mccravyedwin(407)  (Show Source):
You can put this solution on YOUR website!
Here is my approach. In the diagram below, I have placed the number of
forbidden places for the second queen in the event that we choose to put the
first queen in that space. I will explain below, where I got them.
21 21 21 21 21 21 21 21
21 23 23 23 23 23 23 21
21 23 25 25 25 25 23 21
21 23 25 27 27 25 23 21
21 23 25 27 27 25 23 21
21 23 25 25 25 25 23 21
21 23 23 23 23 23 23 21
21 21 21 21 21 21 21 21
Case 1. We put a queen on the edge of the outer 8x8 square.
If we put a queen on the edge of the outer 8x8 square, there are 21
forbidden places for the other queen, 7 on the same row, 7 on the same
column, and 7 on the same diagonal. That's 21 forbidden places. That's
why I put 21's around the outer edge of the 8x8 square and colored them
red. So for each of the 28 places we choose for one queen, there are
63-21=42 safe places to put the other queen. That's (28)(42)=1176.
Case 2. We put a queen on the edge of the inner 6x6 square.
If we put a queen on the edge of the inner 6x6 square, there are 23
forbidden places for the other queen, 7 on the same row, 7 on the same
column, and 9 on the same diagonal, since we pick up 2 more forbidden
places on the diagonals. That's 23 forbidden places. That's
why I put 23's around the edge of the 6x6 square and colored them green.
So for each of the 20 places we choose for one queen, there are 63-23=40
safe places to put the other queen. That's (20)(40)=800.
Case 3. We put a queen on the edge of the inner 4x4 square.
If we put a queen on the edge of the inner 4x4 square, there are 25
forbidden places for the other queen, 7 on the same row, 7 on the same
column, and 11 on the same diagonal, since we pick up 2 more forbidden
places on the diagonals. That's 25 forbidden places. That's
why I put 25's around the edge of the 4x4 square and colored them red.
So for each of the 12 places we choose for one queen, there are 63-25=38
safe places to put the other queen. That's (12)(38)=456.
Case 4. We put a queen on the center 2x2 square.
If we put a queen on the edge of the center 2x2 square, there are 27
forbidden places for the other queen, 7 on the same row, 7 on the same
column, and 13 on the same diagonal, since we pick up 2 more forbidden
places on the diagonals. That's 27 forbidden places. That's
why I put 27's for the center 2x2 square and colored them green.
So for each of the 4 places we choose for one queen, there are 63-27=36
safe places to put the other queen. That's (4)(36)=144.
Total for the 4 cases = 1176 + 800 + 456 + 144 = 2576.
Since we can swap the queens, we take half, and get 1288.
Edwin
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