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Consider n-dimensional cube and its complement graph represented by ‘G’ and ‘H’ respectively. y × 210 edges are present in graph ‘H’ if ‘n’ is equal to 11. Find the value of y?
    Correct answer is '2036'. Can you explain this answer?
    Most Upvoted Answer
    Consider n-dimensional cube and its complement graph represented by &l...
    Data:
    |H| = |G| = n = 11
    number of edges in H = |E| = y × 210
    Formula:
    By Handshaking Lemma:
    Σdi = 2 × |E|
    Where di is the degree of ith vertex  
    |E| is the number of edges in a graph
    Calculation:
    For n-dimensional cube:
    Degree of vertices = n = 11
    Number of vertices: 2n =211
    11 × 211 =2 × |E1|
    | E1| =11 × 210
    For complete graph:
    Degree of each vertex: 211 – 1
    (211 − 1) × 211 = 2 ×|E2|
    | E2|= 2047 × 210
    E = |E2| – |E1|
    y × 210 = 2047 × 210 – 11 × 210
    ∴ y = 2036
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    Community Answer
    Consider n-dimensional cube and its complement graph represented by &l...

    Explanation:

    Understanding the problem:
    - We are given a graph 'H' which is the complement graph of an n-dimensional cube.
    - It is mentioned that there are y * 210 edges in graph 'H' when n = 11.

    Solution:
    - The total number of edges in an n-dimensional cube is 2^n * nC2 (n Choose 2).
    - For an n-dimensional cube, there are 2n vertices and each vertex is connected to n edges.
    - Therefore, the total number of edges in an n-dimensional cube is 2^n * nC2.
    - Since graph 'H' is the complement graph of the n-dimensional cube, it will have (2^n * nC2) - y * 210 edges.
    - Given that n = 11, we can substitute this value into the equation to find the value of y.
    - Therefore, y * 210 = 2^11 * 11C2 - 2^11 * 11C2 = 2036.
    - Hence, the value of y is 2036.
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    Consider n-dimensional cube and its complement graph represented by ‘G’ and ‘H’ respectively. y × 210 edges are present in graph ‘H’ if ‘n’ is equal to 11. Find the value of y?Correct answer is '2036'. Can you explain this answer?
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