I know that a cuboid has 8 vertices and 12 edges, so here's my problem:
I need to evaluate how many scaffold bars are needed to build a cuboid frame; regulations
require that all adjacent vertices pairs have a bar connecting them.
E.g. if the frame was a square ABCD then there would be bars from AB, BC, CD, DA, AC, BD
i.e. four edges and two diagonals.
I know answer is 28 but how!
I need to evaluate how many scaffold bars are needed to build a cuboid frame; regulations
require that all adjacent vertices pairs have a bar connecting them.
E.g. if the frame was a square ABCD then there would be bars from AB, BC, CD, DA, AC, BD
i.e. four edges and two diagonals.
I know answer is 28 but how!
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Consider that each of the 8 vertices is connected to the 7 other vertices by a bar. Counting those 8*7 = 56 connections results in each bar being counted twice (once from each end). Thus, there are
.. 8*7/2 = 4*7 = 28 bars in the frame.
Another way to consider it is that any two vertices are connected by a bar. 8 vertices taken 2 at a time gives 8!/(2!*6!) = 28 different combinations.
Another way to count is
.. 12 edges = 12 bars ... (ab bc cd da a'b' b'c' c'd' d'a' aa' bb' cc' dd')
.. 2 face diagonals * 6 faces = 12 bars ... (ac bd a'c' b'd' ab' ba' bc' cb' cd' dc' da' ad')
.. 4 space diagonals = 4 bars ... (ac' bd' ca' db')
.. 12 + 12 + 4 = 28
.. 8*7/2 = 4*7 = 28 bars in the frame.
Another way to consider it is that any two vertices are connected by a bar. 8 vertices taken 2 at a time gives 8!/(2!*6!) = 28 different combinations.
Another way to count is
.. 12 edges = 12 bars ... (ab bc cd da a'b' b'c' c'd' d'a' aa' bb' cc' dd')
.. 2 face diagonals * 6 faces = 12 bars ... (ac bd a'c' b'd' ab' ba' bc' cb' cd' dc' da' ad')
.. 4 space diagonals = 4 bars ... (ac' bd' ca' db')
.. 12 + 12 + 4 = 28