Answers.Body: <h3>TL;DR: 120</h3>
  <p>You're on the right track, but combinatorics can be hard sometimes.</p>
  <p>If the black king must be in the corner, just pick a1. If the white king is on c1, one knight (the white one below) must give check on c2 or b3, and the other one (the black one below) must cover a2, so be on c3 or b4. That's four positions (we'll cover symmetry later).</p>
  <pre><code>[FEN &quot;8/8/8/8/1n6/1Nn5/2N5/k1K5 w - - 0 1&quot;]
  </code></pre>
  <p>If the white king is on c2, the knight giving (the white one below) check must be on b3 and the other one (the black one below) must cover a2, so be on c1, c3 or b4. That's three positions.</p>
  <pre><code>[FEN &quot;8/8/8/8/1n6/1Nn5/2K5/k1n5 w - - 0 1&quot;]
  </code></pre>
  <p>Now all seven positions can be flipped along the long diagonal; that's 14 (there are no duplicates; the white king isn't on the diagonal). And finally all positions can happen on all four corners, for a total of 14 times 4 = <strong>56</strong> positions ... <strong>in the corner</strong>. @Remellion found another mating pattern:</p>
  <pre><code>[FEN &quot;8/8/8/8/8/1N2N3/K7/K1k5 w - - 0 1&quot;]
  </code></pre>
  <p>(the white king can be at two places) and this position can be shifted one square to the right four times. (I missed it because I only looked for mates with the black king on b1 and assumed I didn't need to consider the other edge squares. I was wrong ...) We again have an eight-fold rotation symmetry (four edges times two, for the mirrored version), so this pattern yield 4 times 2 times 8 = <strong>64</strong> positions.</p>
  <h3>Together, there are 120 mating positions.</h3>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-04-07T02:24:18.990Z
Answers.Text: It is possible to also mate on the edge. This answer is incomplete. For instance: 8/8/8/8/8/2N2N2/1K6/3k4 w - -
Answers.UserDisplayName: Remellion
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-04-06T19:07:24.173Z
Answers.LastActivityDate: 2019-04-07T07:06:23.710Z
Answers.LastEditDate: 2020-06-17T09:59:59.497Z
Answers.UserDisplayName: Glorfindel
Body: <p>What is the total number of KNN vs. K checkmate positions? Assume that White is checkmating.</p>
  
  <p>Firstly the Black king must be mated on a corner. <strong>It is impossible to do it on the edge.</strong> The calculations on one corner can be multiplied by four for the final results.</p>
  
  <p><strong>Addendum</strong>: It actually can be done as @Glorfindel’s answer shows. </p>
  
  <pre><code>[FEN "k6k/8/8/8/8/8/8/k6k w - - 0 1"]
  </code></pre>
  
  <p>Let’s choose this corner for now.</p>
  
  <pre><code>[FEN "8/8/8/8/8/8/8/k7 w - - 0 1"]
  </code></pre>
  
  <p>The white king needs to be on one of these four squares.</p>
  
  <pre><code>[FEN "k6k/8/8/8/8/8/8/k6k w - - 0 1"]
  </code></pre>
  
  <p>The checkmating knight needs to be on one of these two squares.</p>
  
  <pre><code>[FEN "8/8/8/8/8/1N6/2N5/8 w - - 0 1"]
  </code></pre>
  
  <p>Finally, the knight used to block off the black king’s escape route has to be on one of these five squares.</p>
  
  <pre><code>[FEN "8/8/8/8/1N6/N1N5/3N4/2N5 w - - 0 1"]
  </code></pre>
  
  <p>Can anyone calculate the total number of possiblities?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2019-04-07T02:23:55.460Z
Comments.Text: You are making a wrong assumption: You can mate on the edge. For instance: 8/8/8/8/8/2N2N2/1K6/3k4 w - - I count another 4x2x8 = 64 (4 squares on the edge, 2 positions of white king, multiplicity 8 by rotation and reflection symmetry) positions.
Comments.UserDisplayName: Remellion
Tags: <checkmate><knights><mathematics><knight-endgame>
Title: What is the total number of KNN vs K checkmate positions?
ThreadText: # What is the total number of KNN vs K checkmate positions?
  
  (Asked by: username_0 on 2019-04-06)
  
  What is the total number of KNN vs. K checkmate positions? Assume that White is checkmating.
  
  Firstly the Black king must be mated on a corner. **It is impossible to do it on the edge.** The calculations on one corner can be multiplied by four for the final results.
  
  **Addendum**: It actually can be done as @username_2’s answer shows.
  
  ```
  [FEN "k6k/8/8/8/8/8/8/k6k w - - 0 1"]
  
  ```
  
  Let’s choose this corner for now.
  
  ```
  [FEN "8/8/8/8/8/8/8/k7 w - - 0 1"]
  
  ```
  
  The white king needs to be on one of these four squares.
  
  ```
  [FEN "k6k/8/8/8/8/8/8/k6k w - - 0 1"]
  
  ```
  
  The checkmating knight needs to be on one of these two squares.
  
  ```
  [FEN "8/8/8/8/8/1N6/2N5/8 w - - 0 1"]
  
  ```
  
  Finally, the knight used to block off the black king’s escape route has to be on one of these five squares.
  
  ```
  [FEN "8/8/8/8/1N6/N1N5/3N4/2N5 w - - 0 1"]
  
  ```
  
  Can anyone calculate the total number of possiblities?
  
  **Tags:** checkmate, knights, mathematics, knight-endgame
  
  > **Comments:**
  > **username_1**: You are making a wrong assumption: You can mate on the edge. For instance: 8/8/8/8/8/2N2N2/1K6/3k4 w - - I count another 4x2x8 = 64 (4 squares on the edge, 2 positions of white king, multiplicity 8 by rotation and reflection symmetry) positions.  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_2 on 2019-04-06. Score: 3 | Accepted answer
  
  ### TL;DR: 120
  
  You're on the right track, but combinatorics can be hard sometimes.
  
  If the black king must be in the corner, just pick a1. If the white king is on c1, one knight (the white one below) must give check on c2 or b3, and the other one (the black one below) must cover a2, so be on c3 or b4. That's four positions (we'll cover symmetry later).
  
  ```
  [FEN "8/8/8/8/1n6/1Nn5/2N5/k1K5 w - - 0 1"]
  
  ```
  
  If the white king is on c2, the knight giving (the white one below) check must be on b3 and the other one (the black one below) must cover a2, so be on c1, c3 or b4. That's three positions.
  
  ```
  [FEN "8/8/8/8/1n6/1Nn5/2K5/k1n5 w - - 0 1"]
  
  ```
  
  Now all seven positions can be flipped along the long diagonal; that's 14 (there are no duplicates; the white king isn't on the diagonal). And finally all positions can happen on all four corners, for a total of 14 times 4 = **56** positions ... **in the corner**. @username_1 found another mating pattern:
  
  ```
  [FEN "8/8/8/8/8/1N2N3/K7/K1k5 w - - 0 1"]
  
  ```
  
  (the white king can be at two places) and this position can be shifted one square to the right four times. (I missed it because I only looked for mates with the black king on b1 and assumed I didn't need to consider the other edge squares. I was wrong ...) We again have an eight-fold rotation symmetry (four edges times two, for the mirrored version), so this pattern yield 4 times 2 times 8 = **64** positions.
  
  ### Together, there are 120 mating positions.
  
  > **Comments:**
  > **username_1**: It is possible to also mate on the edge. This answer is incomplete. For instance: 8/8/8/8/8/2N2N2/1K6/3k4 w - -  
  
  ---
  
  