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Question: There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. Th...

There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through at least three points of these points is
(a) 116
(b) 120
(c) 117
(d) none of these

Explanation

Solution

We will solve this question with the help of combinations and hence we will use the formula nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}. Non collinear means points are not in a line and are random whereas concyclic means if we draw the circle using three points it should pass from the 4th point.

Complete step-by-step answer:
Now it is mentioned in the question that 3 points are not collinear and 4 points are concyclic and the total number of points given is 10.
So we can get circles from joining 3 non collinear points out of 10 but this also includes circles formed from joining 3 points out of 4 concyclic points and hence we will subtract these circles because they are repetitive. And now we will add 1 as there is a circle which would pass all 4 points. Hence using this information, we get,
Total number of circles =10C3−4C3+1.......(1)={}^{10}{{C}_{3}}-{}^{4}{{C}_{3}}+1.......(1)
Now applying the formula nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!} in equation (1), we get,
⇒10!3!(10−3)!−4!3!(4−3)!+1.......(2)\Rightarrow \dfrac{10!}{3!(10-3)!}-\dfrac{4!}{3!(4-3)!}+1.......(2)
Now expanding the factorials and solving equation (2) we get,
⇒10×9×8×7!3×2×7!−4×3!3!+1.......(3)\Rightarrow \dfrac{10\times 9\times 8\times 7!}{3\times 2\times 7!}-\dfrac{4\times 3!}{3!}+1.......(3)
Now cancelling similar terms in equation (3), we get,

& \Rightarrow \dfrac{10\times 9\times 8}{3\times 2}-4+1 \\\ & \Rightarrow 10\times 3\times 4-4+1=120-4+1=117 \\\ \end{aligned}$$ **Hence the number of different circles is 117. So the correct answer is option (c).** **Note:** A combination is a way to order or arrange a set or number of things uniquely. And knowing this formula $${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$$ is important . We can make a mistake in solving equation (1) if we do not properly expand $${}^{10}{{C}_{3}}$$ and $${}^{4}{{C}_{3}}$$. Also understanding the concept of collinear and concyclic is important.