Cayley's formula and Graph theory












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We have 3 spanning trees. First consisting of 2 vertices, second from 1 vertice and third from 1 vertice. Total number of nodes in the graph is 4. We have to connect these three spanning trees to form one. On how many different ways we can do this. I know we have to use Cayley's Formula but I do not know how



example: 4 {1} {2} Returns: 8 There are eight spanning trees that contain the edge
(1, 2):
{(1, 2), (1, 3), (1, 4)}
{(1, 2), (1, 3), (2, 4)}
{(1, 2), (1, 3), (3, 4)}
{(1, 2), (2, 3), (2, 4)}
{(1, 2), (1, 4), (2, 3)}
{(1, 2), (1, 4), (3, 4)}
{(1, 2), (2, 3), (3, 4)}
{(1, 2), (2, 4), (3, 4)}



We have 4 vertices, two of them are already connected (1,2) and 3 and 4 are disconnected. we need to connect 3 and 4 to (1,2). I need an explanation of this formula: (n)^(k – 2) * s_0 * s_1 * … * s_{k-1} which confirms the above example n-number of vertices k-connected component










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    $begingroup$


    We have 3 spanning trees. First consisting of 2 vertices, second from 1 vertice and third from 1 vertice. Total number of nodes in the graph is 4. We have to connect these three spanning trees to form one. On how many different ways we can do this. I know we have to use Cayley's Formula but I do not know how



    example: 4 {1} {2} Returns: 8 There are eight spanning trees that contain the edge
    (1, 2):
    {(1, 2), (1, 3), (1, 4)}
    {(1, 2), (1, 3), (2, 4)}
    {(1, 2), (1, 3), (3, 4)}
    {(1, 2), (2, 3), (2, 4)}
    {(1, 2), (1, 4), (2, 3)}
    {(1, 2), (1, 4), (3, 4)}
    {(1, 2), (2, 3), (3, 4)}
    {(1, 2), (2, 4), (3, 4)}



    We have 4 vertices, two of them are already connected (1,2) and 3 and 4 are disconnected. we need to connect 3 and 4 to (1,2). I need an explanation of this formula: (n)^(k – 2) * s_0 * s_1 * … * s_{k-1} which confirms the above example n-number of vertices k-connected component










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      We have 3 spanning trees. First consisting of 2 vertices, second from 1 vertice and third from 1 vertice. Total number of nodes in the graph is 4. We have to connect these three spanning trees to form one. On how many different ways we can do this. I know we have to use Cayley's Formula but I do not know how



      example: 4 {1} {2} Returns: 8 There are eight spanning trees that contain the edge
      (1, 2):
      {(1, 2), (1, 3), (1, 4)}
      {(1, 2), (1, 3), (2, 4)}
      {(1, 2), (1, 3), (3, 4)}
      {(1, 2), (2, 3), (2, 4)}
      {(1, 2), (1, 4), (2, 3)}
      {(1, 2), (1, 4), (3, 4)}
      {(1, 2), (2, 3), (3, 4)}
      {(1, 2), (2, 4), (3, 4)}



      We have 4 vertices, two of them are already connected (1,2) and 3 and 4 are disconnected. we need to connect 3 and 4 to (1,2). I need an explanation of this formula: (n)^(k – 2) * s_0 * s_1 * … * s_{k-1} which confirms the above example n-number of vertices k-connected component










      share|cite|improve this question









      $endgroup$




      We have 3 spanning trees. First consisting of 2 vertices, second from 1 vertice and third from 1 vertice. Total number of nodes in the graph is 4. We have to connect these three spanning trees to form one. On how many different ways we can do this. I know we have to use Cayley's Formula but I do not know how



      example: 4 {1} {2} Returns: 8 There are eight spanning trees that contain the edge
      (1, 2):
      {(1, 2), (1, 3), (1, 4)}
      {(1, 2), (1, 3), (2, 4)}
      {(1, 2), (1, 3), (3, 4)}
      {(1, 2), (2, 3), (2, 4)}
      {(1, 2), (1, 4), (2, 3)}
      {(1, 2), (1, 4), (3, 4)}
      {(1, 2), (2, 3), (3, 4)}
      {(1, 2), (2, 4), (3, 4)}



      We have 4 vertices, two of them are already connected (1,2) and 3 and 4 are disconnected. we need to connect 3 and 4 to (1,2). I need an explanation of this formula: (n)^(k – 2) * s_0 * s_1 * … * s_{k-1} which confirms the above example n-number of vertices k-connected component







      graph-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 16 at 16:30









      Kristian ČotićKristian Čotić

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