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How many symmetric relations?

  1. May 9, 2010 #1

    Let A = 1,2,3,4,5,6,7
    How many symmetric relations on A contain exactly (a) four ordered pairs, (b) 5 , (c) seven and (d) eight

    The book has solutions to the first two, which I didn't understand at all.

    Please look the pic below

    Can someone guide me through how to approach the problem?

  2. jcsd
  3. May 11, 2010 #2
    You can imagine any relation a a table with 7 rows and 7 columns, where you indicate in each position whether the corresponding elements belong to the relation. (This is basically the same as working with the graph of this relation.)

    A relation is symmetric if and only if this graph/table is symmetric with respect to the main diagonal. Hence, a symmetric relation is uniquely determined by the pairs on and above the main diagonal.
    You have 7 positions on diagonal and 21=6+5+4+3+2+1 positions above the diagonal.
    If you put a elements above the diagonal, then there are also a elements bellow it, by the symmetry. So, by putting a elements above the diagonal and b elements on the diagonal, you obtain a relation consisting of 2a+b pairs.

    Now, if you want to have 4 elements you have these possibilities:
    all 4 elements are on the diagonal [tex]\binom{7}{4}[/tex];
    2 elements on the diagonal and 1 above it [tex]\binom72\binom{21}1[/tex];
    no element on the diagonal and 2 above it [tex]\binom{21}2[/tex].

    For 5 elements you have the following possibilieites:
    All 5 elements on the diagonal [tex]\binom75[/tex];
    3 elements on the diagonal and 1 above it [tex]\binom73+\binom{21}1[/tex];
    1 element on the diagonal and 2 above it [tex]\binom71+\binom{21}2[/tex].
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