Question

The number of triangles that can be formed with 10 points as vertices, n of them being collinear, is 110. Then n is
(a) 3
(b) 4
(c) 5
(d) 6

Answer 1

We will solve this question with the help of combinations and hence we will use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$. Collinear points means that they are in a straight line. So a triangle cannot be formed from collinear points.

Complete step-by-step answer:
It is given in the question that n points out of 10 points are collinear. And so we cannot form a triangle from the n collinear points.
So if we select any 3 points out of the 10 points we can form triangles and the total number of triangles formed is 110. Also if we select 3 points from the n collinear points we cannot form triangles. So using this information we get,
$\Rightarrow {}^{10}{{C}_{3}}-{}^{n}{{C}_{3}}=110........(1)$
Now applying the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ in equation (1), we get,
$\Rightarrow \dfrac{10!}{3!(10-3)!}-\dfrac{n!}{3!(n-3)!}=110........(2)$
Now expanding the factorials in equation (2) we get,
$\Rightarrow \dfrac{10\times 9\times 8\times 7!}{3\times 2\times 7!}-\dfrac{n(n-1)(n-2)(n-3)!}{3\times 2\times (n-3)!}=110........(3)$
Now cancelling the similar terms in equation (3) we get,

& \Rightarrow \dfrac{10\times 9\times 8}{3\times 2}-\dfrac{n(n-1)(n-2)}{3\times 2}=110 \\\ & \Rightarrow \dfrac{720}{6}-\dfrac{n(n-1)(n-2)}{6}=110...........(4) \\\ \end{aligned}$$ Now simplifying the terms in equation (4) we get, $$\Rightarrow 120-\dfrac{n(n-1)(n-2)}{6}=110...........(5)$$ So from equation (5) we can see that the second term in the left hand side should be equal to 10 so that the right hand side becomes equal to 110. Hence we get, $$\begin{aligned} & \Rightarrow n(n-1)(n-2)=2\times 5\times 3\times 2 \\\ & \Rightarrow n(n-1)(n-2)=5\times 4\times 3......(6) \\\ \end{aligned}$$ So from equation (6) we can see that the product is of consecutive integers in the left hand side and also in the right hand side and hence n is 5. **So the correct answer is option (c).** **Note:** Remembering the concept of collinear points and combinations is the key here. A combination is a way to order or arrange a set or number of things uniquely. We can make a mistake in solving equation (1) if we do not properly expand $${}^{10}{{C}_{3}}$$ and $${}^{n}{{C}_{3}}$$. Also if we would have expanded the terms by multiplying in equation (5) we would have got a cubic polynomial and hence to avoid this we solved it using the concept of consecutive integers.