{"@context":"https://schema.org/","@type":"Quiz","dateCreated":"2022-07-07T09:58:59.917Z","dateModified":"2022-07-07T09:58:59.917Z","typicalAgeRange":"11-18","educationalLevel":"","assesses":"Mathematics - permutations and combinations","educationalAlignment":[{"@type":"AlignmentObject","alignmentType":"educationalTopic","targetName":"permutations and combinations"},{"@type":"AlignmentObject","alignmentType":"educationalSubject","targetName":"Mathematics"},{"@type":"AlignmentObject","alignmentType":"educationalExam","targetName":""}],"name":"Quiz about permutations and combinations","about":{"@type":"Thing","name":"Mathematics - permutations and combinations"},"hasPart":[{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"intermediate","name":"Question About permutations and combinations","text":"There are $10$ points in a plane of which $4$ are collinear. The number of quadrilaterals that can be formed is}","encodingFormat":"text/markdown","comment":{"@type":"Comment","text":"\"\""},"suggestedAnswer":[{"@type":"Answer","position":0,"encodingFormat":"text/markdown","text":"

$15$

","comment":{"@type":"Comment","text":"\"\""}},{"@type":"Answer","position":1,"encodingFormat":"text/markdown","text":"

$45$

","comment":{"@type":"Comment","text":"\"\""}},{"@type":"Answer","position":2,"encodingFormat":"text/markdown","text":"

$50$

","comment":{"@type":"Comment","text":"\"\""}},{"@type":"Answer","position":3,"encodingFormat":"text/markdown","text":"

$185$

","comment":{"@type":"Comment","text":"\"\""}}],"acceptedAnswer":{"@type":"Answer","position":3,"encodingFormat":"text/markdown","text":"

$185$

","comment":{"@type":"Comment","text":"\"\""},"answerExplanation":{"@type":"comment","text":"To form a quadrilateral, $4$ points are required. They can be (i) All the $4$ points from $10 - 4 = 6$ non-collinear points. (ii) $3$ points from $6$ non-collinear points and $1$ point from $4$ collinear points. (iii) $2$ points from $6$ non-collinear points and $2$ points from $4$ collinear points. $\\therefore$ reqd. no. of ways $=\\,^{6}C_{4}+\\,^{6}C_{3}\\cdot\\,^{4}C_{1}+\\,^{6}C_{2}\\cdot\\,^{4}C_{2}$ $=\\frac{6\\times5}{1\\times2}+\\frac{6\\times5\\times4}{1\\times2\\times3}\\times4+\\frac{6\\times5}{1\\times2}\\cdot\\frac{4\\times3}{1\\times2}$ $= 15 + 80 + 90$ $=185$"}}}]}

There are $10$ points in a plane of which $4$ are collinear. The number of quadrilaterals that can be formed is

The Correct Option is D

Solution and Explanation

Concepts Used:

Permutations and Combinations

Permutation:

Combination: