Let there be $16$ points in a plane out of which $6$ lie on one line and $5$ lie on other line. Then the correct option(s) among the following is/are

Total straight lines that can be formed using these points are

95

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Total triangles that can be formed using these points are

530

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Total circles that can be formed using these points are

256

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Total quadrilaterals that can be formed using these points are

1490

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Solution

The correct option is D Total quadrilaterals that can be formed using these points are $1490$
Total straight lines that can be formed from $16$ points $=^{16} C_{2}$
Since total points are $16$ , but $6$ are collinear and other $5$ are also collinear,
$∴$ Required number of ways $=^{16} C_{2} -^{6} C_{2} -^{5} C_{2} + 1 + 1 = 97$
(two lines will be obtained from collinear points)

Total triangles that can be formed from $16$ points $=^{16} C_{3}$
Since total points are $16$ , but $6$ are collinear and other $5$ are also collinear,
$∴$ Required number of ways $=^{16} C_{3} -^{6} C_{3} -^{5} C_{3} = 530$

Total circles that can be formed from $16$ points $=^{16} C_{3}$
Since total points are $16$ , but $6$ are collinear and other $5$ are also collinear,
$∴$ Required number of ways $=^{16} C_{3} -^{6} C_{3} -^{5} C_{3} = 530$

Total quadrilaterals that can be formed from $16$ points $=^{16} C_{4}$
Total points are $16$ , but $6$ are collinear and other $5$ are also collinear,
Also, for making quadrilaterals three collinear points can't be taken.
$∴$ Required number of ways $=^{16} C_{4} -^{6} C_{4} -^{5} C_{4} -^{6} C_{3} \times^{10} C_{1} -^{5} C_{3} \times^{11} C_{1} = 1490$

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Combination

Standard XI Mathematics

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Let there be 16 points in a plane out of which 6 lie on one line and 5 lie on other line. Then the correct option(s) among the following is/are

Let there be $16$ points in a plane out of which $6$ lie on one line and $5$ lie on other line. Then the correct option(s) among the following is/are