<p>There are an even number of points in a plane of which half are lying on the same line and no other three are collinear. If the total number of triangles that can be made with vertices at these points is 110, then the number of given points is</p>
Step-by-Step Solution
Key Concept: When n points include k collinear points, the number of triangles formed is C(n,3) minus the combinations of 3 points chosen entirely from the collinear set: C(n,3) - C(k,3). Set this equal to 110 and solve for n.
<p><strong>Step 1:</strong> Let the total number of points be n (even). Then n/2 points are collinear.</p><p><strong>Step 2:</strong> Total ways to choose 3 points from n points = C(n,3) = n(n-1)(n-2)/6</p><p><strong>Step 3:</strong> Degenerate cases (3 collinear points) = C(n/2, 3) = (n/2)(n/2-1)(n/2-2)/6</p><p><strong>Step 4:</strong> Number of valid triangles = C(n,3) - C(n/2, 3) = 110</p><p><strong>Step 5:</strong> n(n-1)(n-2)/6 - (n/2)(n/2-1)(n/2-2)/6 = 110</p><p><strong>Step 6:</strong> Simplify: [n(n-1)(n-2) - (n/2)(n/2-1)(n/2-2)]/6 = 110</p><p><strong>Step 7:</strong> Multiply by 6: n(n-1)(n-2) - (n/2)(n/2-1)(n/2-2) = 660</p><p><strong>Step 8:</strong> Let n = 2m: 2m(2m-1)(2m-2) - m(m-1)(m-2) = 660</p><p><strong>Step 9:</strong> 4m(2m-1)(m-1) - m(m-1)(m-2) = 660</p><p><strong>Step 10:</strong> m(m-1)[4(2m-1) - (m-2)] = 660</p><p><strong>Step 11:</strong> m(m-1)[8m - 4 - m + 2] = 660</p><p><strong>Step 12:</strong> m(m-1)(7m - 2) = 660</p><p><strong>Step 13:</strong> Testing m = 5: 5(4)(33) = 660 ✓</p><p><strong>Step 14:</strong> Therefore n = 2m = 10</p><p>∴ Answer: B (10 points)</p>
Correct Answer: B