How many different (mutually non-congruent) trapeziums can be constructed using four distinct side lengths from the set $\{1, 2, 3, 4, 5, 6\}$?

  • A
    $5$
  • B
    $11$
  • C
    $15$
  • D
    $30$

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In a plane,there are $10$ points,out of which $4$ are collinear. The number of triangles that can be formed by joining these points is:

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