Question

There are n points in a plane out of these points no three are in the same straight line except p points which are collinear. Let "k" be the number of straight lines and "m" be the number of triangles .Then find m−k ?(Assume n=7p=5)

Answer 1

Hint: To find m−k , we have to find the required number of straight lines and triangles. We know that for a line to be formed, 2 points are required. Hence, Number of lines formed with n points ${{=}^{n}}{{C}_{2}}$ . Also a number of lines formed with p collinear points ${{=}^{p}}{{C}_{2}}$ . Hence, required number of straight lines $=k{{=}^{n}}{{C}_{2}}{{-}^{p}}{{C}_{2}}+1$ . We know that a triangle has 3 vertices. Hence, number of triangles formed from n points ${{=}^{n}}{{C}_{3}}$ and number of triangles formed from p points ${{=}^{p}}{{C}_{3}}$ . Hence, required number of triangles $=m{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}$ . Now, we can find m-k by substituting n=5 and $p=\dfrac{5}{7}$ in $m-k{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}-\left( ^{n}{{C}_{2}}{{-}^{p}}{{C}_{2}}+1 \right)$ .

Complete step-by-step answer:
We are given that there are n points in a plane of which no three are in the same straight line except p points which are collinear. Let us find the number of lines formed with n points.
We know that for a line to be formed, 2 points are required. Hence,
Number of lines formed with n points ${{=}^{n}}{{C}_{2}}$
Now, let us find the number of lines formed by collinear points.
Number of lines formed with p collinear points ${{=}^{p}}{{C}_{2}}$
We are given that no three points are in the same straight line except p points which are collinear. The collinear points form one straight line. Hence, we can find the required number of straight lines as follows.
Required number of straight lines ${{=}^{n}}{{C}_{2}}{{-}^{p}}{{C}_{2}}+1$
We added 1 as the collinear points form one straight line.
We are given that the number of straight lines are k. Hence,
$k{{=}^{n}}{{C}_{2}}{{-}^{p}}{{C}_{2}}+1...(i)$
We know that a triangle has 3 vertices. Hence,
Number of triangles formed from n points ${{=}^{n}}{{C}_{3}}$
Number of triangles formed from p points ${{=}^{p}}{{C}_{3}}$
Now we can find the number of triangles formed with n points but not with collinear points.
Required number of triangles ${{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}$
We are given that m is the number of triangles formed. Hence,
$m{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}$
We have to find $m-k$ .
$\Rightarrow m-k{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}-\left( ^{n}{{C}_{2}}{{-}^{p}}{{C}_{2}}+1 \right)$
We can write this as
$m-k{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}{{-}^{n}}{{C}_{2}}{{+}^{p}}{{C}_{2}}-1$
We are given that $n=5$ .
$\Rightarrow m-k{{=}^{5}}{{C}_{3}}{{-}^{p}}{{C}_{3}}{{-}^{5}}{{C}_{2}}{{+}^{p}}{{C}_{2}}-1$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Let’s expand the above equation.
$\begin{align}
  & \Rightarrow m-k=\dfrac{5!}{3!\left( 5-3 \right)!}{{-}^{p}}{{C}_{3}}-\dfrac{5!}{2!\left( 5-2 \right)!}{{+}^{p}}{{C}_{2}}-1 \\
 & \Rightarrow m-k=\dfrac{5!}{3!2!}{{-}^{p}}{{C}_{3}}-\dfrac{5!}{2!3!}{{+}^{p}}{{C}_{2}}-1 \\
\end{align}$
Let us cancel the common terms. We will get
$\Rightarrow m-k={{-}^{p}}{{C}_{3}}{{+}^{p}}{{C}_{2}}-1$
Now, let’s again apply $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to the above equation. We will get
$\Rightarrow m-k=-\dfrac{p!}{3!\left( p-3 \right)!}+\dfrac{p!}{2!\left( p-2 \right)!}-1$
We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times ...1$ . Hence,
$\Rightarrow m-k=-\dfrac{p\left( p-1 \right)\left( p-2 \right)\left( p-3 \right)!}{3!\left( p-3 \right)!}+\dfrac{p\left( p-1 \right)\left( p-2 \right)!}{2!\left( p-2 \right)!}-1$
Let us cancel the common terms. We will get
$m-k=-\dfrac{p\left( p-1 \right)\left( p-2 \right)}{3!}+\dfrac{p\left( p-1 \right)}{2!}-1$
Now, let’s expand the factorial.
$\begin{align}
  & \Rightarrow m-k=-\dfrac{p\left( p-1 \right)\left( p-2 \right)}{3\times 2\times 1}+\dfrac{p\left( p-1 \right)}{2\times 1}-1 \\
 & \Rightarrow m-k=-\dfrac{p\left( p-1 \right)\left( p-2 \right)}{6}+\dfrac{p\left( p-1 \right)}{2}-1 \\
\end{align}$
Let us take the LCM of 6 and 2 which is 6. We can write the above equation as
$\Rightarrow m-k=\dfrac{-p\left( p-1 \right)\left( p-2 \right)+3p\left( p-1 \right)-6}{6}$
Let us simplify this.
$\begin{align}
  & \Rightarrow m-k=\dfrac{-\left( {{p}^{2}}-p \right)\left( p-2 \right)+3{{p}^{2}}-3p-6}{6} \\
 & \Rightarrow m-k=\dfrac{-{{p}^{3}}+2{{p}^{2}}+{{p}^{2}}-2p+3{{p}^{2}}-3p-6}{6} \\
\end{align}$
Let’s solve this.
$\Rightarrow m-k=\dfrac{-{{p}^{3}}+6{{p}^{2}}-5p-6}{6}$
We are given that $7p=5\Rightarrow p=\dfrac{5}{7}$ . Let’s substitute this in the above equation.
$\Rightarrow m-k=\dfrac{-{{\left( \dfrac{5}{7} \right)}^{3}}+6{{\left( \dfrac{5}{7} \right)}^{2}}-5\left( \dfrac{5}{7} \right)-6}{6}$
Let’s expand this.
$\begin{align}
  & \Rightarrow m-k=\dfrac{-\dfrac{125}{343}+6\times \dfrac{25}{49}-\dfrac{25}{7}-6}{6} \\
 & \Rightarrow m-k=\dfrac{-\dfrac{125}{343}+\dfrac{150}{49}-\dfrac{25}{7}-6}{6} \\
\end{align}$
Let us take the LCM of 343,49 and 7 which is 343. We will get

$\begin{align}
  & \Rightarrow m-k=\dfrac{\dfrac{-125+1050-1225-2058}{343}}{6} \\
 & \Rightarrow m-k=\dfrac{\dfrac{-125+1050-1225-2058}{343}}{6} \\
\end{align}$
Let’s solve this. We will get
$\begin{align}
  & \Rightarrow m-k=\dfrac{-2358}{343\times 6} \\
 & \Rightarrow m-k=\dfrac{-2358}{2058}=\dfrac{-393}{343} \\
\end{align}$
Let us write this in decimals. We will get
$m-k=1.145$
Hence, the value of $m-k$ is 1.145.

Note: You may make mistake by writing required number of straight lines ${{=}^{n}}{{C}_{2}}{{-}^{p}}{{C}_{2}}-1$ and also by writing required number of triangles ${{=}^{n}}{{C}_{3}}{{+}^{p}}{{C}_{3}}$ . Do not substitute the value of p at the beginning as we will not be able to solve it further. You must find the value of p from $7p=5$ before substituting.