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Question: There are n points in a plane out of these points no three are in the same straight line except p po...

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)

Explanation

Solution

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 =nC2{{=}^{n}}{{C}_{2}} . Also a number of lines formed with p collinear points =pC2{{=}^{p}}{{C}_{2}} . Hence, required number of straight lines =k=nC2−pC2+1=k{{=}^{n}}{{C}_{2}}{{-}^{p}}{{C}_{2}}+1 . We know that a triangle has 3 vertices. Hence, number of triangles formed from n points =nC3{{=}^{n}}{{C}_{3}} and number of triangles formed from p points =pC3{{=}^{p}}{{C}_{3}} . Hence, required number of triangles =m=nC3−pC3=m{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}} . Now, we can find m-k by substituting n=5 and p=57p=\dfrac{5}{7} in m−k=nC3−pC3−(nC2−pC2+1)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 =nC2{{=}^{n}}{{C}_{2}}
Now, let us find the number of lines formed by collinear points.
Number of lines formed with p collinear points =pC2{{=}^{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 =nC2−pC2+1{{=}^{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=nC2−pC2+1...(i)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 =nC3{{=}^{n}}{{C}_{3}}
Number of triangles formed from p points =pC3{{=}^{p}}{{C}_{3}}
Now we can find the number of triangles formed with n points but not with collinear points.
Required number of triangles =nC3−pC3{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}
We are given that m is the number of triangles formed. Hence,
m=nC3−pC3m{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}
We have to find m−km-k .
⇒m−k=nC3−pC3−(nC2−pC2+1)\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=nC3−pC3−nC2+pC2−1m-k{{=}^{n}}{{C}_{3}}{{-}^{p}}{{C}_{3}}{{-}^{n}}{{C}_{2}}{{+}^{p}}{{C}_{2}}-1
We are given that n=5n=5 .
⇒m−k=5C3−pC3−5C2+pC2−1\Rightarrow m-k{{=}^{5}}{{C}_{3}}{{-}^{p}}{{C}_{3}}{{-}^{5}}{{C}_{2}}{{+}^{p}}{{C}_{2}}-1
We know that nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} . Let’s expand the above equation.
⇒m−k=5!3!(5−3)!−pC3−5!2!(5−2)!+pC2−1 ⇒m−k=5!3!2!−pC3−5!2!3!+pC2−1 \begin{aligned} & \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{aligned}
Let us cancel the common terms. We will get
⇒m−k=−pC3+pC2−1\Rightarrow m-k={{-}^{p}}{{C}_{3}}{{+}^{p}}{{C}_{2}}-1
Now, let’s again apply nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} to the above equation. We will get
⇒m−k=−p!3!(p−3)!+p!2!(p−2)!−1\Rightarrow m-k=-\dfrac{p!}{3!\left( p-3 \right)!}+\dfrac{p!}{2!\left( p-2 \right)!}-1
We know that n!=n×(n−1)×(n−2)×(n−3)×...1n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times ...1 . Hence,
⇒m−k=−p(p−1)(p−2)(p−3)!3!(p−3)!+p(p−1)(p−2)!2!(p−2)!−1\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=−p(p−1)(p−2)3!+p(p−1)2!−1m-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.
⇒m−k=−p(p−1)(p−2)3×2×1+p(p−1)2×1−1 ⇒m−k=−p(p−1)(p−2)6+p(p−1)2−1 \begin{aligned} & \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{aligned}
Let us take the LCM of 6 and 2 which is 6. We can write the above equation as
⇒m−k=−p(p−1)(p−2)+3p(p−1)−66\Rightarrow m-k=\dfrac{-p\left( p-1 \right)\left( p-2 \right)+3p\left( p-1 \right)-6}{6}
Let us simplify this.
⇒m−k=−(p2−p)(p−2)+3p2−3p−66 ⇒m−k=−p3+2p2+p2−2p+3p2−3p−66 \begin{aligned} & \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{aligned}
Let’s solve this.
⇒m−k=−p3+6p2−5p−66\Rightarrow m-k=\dfrac{-{{p}^{3}}+6{{p}^{2}}-5p-6}{6}
We are given that 7p=5⇒p=577p=5\Rightarrow p=\dfrac{5}{7} . Let’s substitute this in the above equation.
⇒m−k=−(57)3+6(57)2−5(57)−66\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.
⇒m−k=−125343+6×2549−257−66 ⇒m−k=−125343+15049−257−66 \begin{aligned} & \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{aligned}
Let us take the LCM of 343,49 and 7 which is 343. We will get

⇒m−k=−125+1050−1225−20583436 ⇒m−k=−125+1050−1225−20583436 \begin{aligned} & \Rightarrow m-k=\dfrac{\dfrac{-125+1050-1225-2058}{343}}{6} \\\ & \Rightarrow m-k=\dfrac{\dfrac{-125+1050-1225-2058}{343}}{6} \\\ \end{aligned}
Let’s solve this. We will get
⇒m−k=−2358343×6 ⇒m−k=−23582058=−393343 \begin{aligned} & \Rightarrow m-k=\dfrac{-2358}{343\times 6} \\\ & \Rightarrow m-k=\dfrac{-2358}{2058}=\dfrac{-393}{343} \\\ \end{aligned}
Let us write this in decimals. We will get
m−k=1.145m-k=1.145
Hence, the value of m−km-k is 1.145.

Note: You may make mistake by writing required number of straight lines =nC2−pC2−1{{=}^{n}}{{C}_{2}}{{-}^{p}}{{C}_{2}}-1 and also by writing required number of triangles =nC3+pC3{{=}^{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=57p=5 before substituting.