Question

If PQRS is a convex quadrilateral with 3, 4, 5 and 6 points marked on side PQ, QR, RS and PS respectively. Then the number of triangles with vertices on different sides is:
A) 220
B) 270
C) 282
D) 342

Answer 1

Hint: Number of sides of a triangle is 3; consider 3 sides of the convex quadrilateral to bring out the number of triangles.

Complete step by step answer:
A convex quadrilateral is a 4 sided polygon that has interior angles that measures less than 180 degrees each.
Consider the convex quadrilateral PQRS shown in the figure.

\[\begin{align}
  & \therefore PQ={}^{3}{{C}_{1}} \\
 & QR={}^{4}{{C}_{1}} \\
 & RS={}^{5}{{C}_{1}} \\
 & PS={}^{6}{{C}_{1}} \\
\end{align}\]

3, 4, 5 and 6 points are marked on sides PQ, QR, RS and PS. Now choose one point from the points from the three points on \[PQ\Rightarrow {}^{3}{{C}_{1}}\]. We need to find the number of triangles with vertices on different sides.

Now, from the figure,

The number of triangle with vertices on sides PQ, QR, RS\[={}^{3}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{5}{{C}_{1}}\]

Number of triangle with vertices on sides QR, SR, SP\[={}^{3}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{5}{{C}_{1}}\]

Number of triangle with vertices on sides RS, PS, PQ \[={}^{5}{{C}_{1}}\times {}^{6}{{C}_{1}}\times {}^{3}{{C}_{1}}\]

Number of triangle with vertices on sides PS, PQ, QR \[={}^{6}{{C}_{1}}\times {}^{3}{{C}_{1}}\times {}^{4}{{C}_{1}}\]

\[\therefore \] Total number of triangles= number of triangle with vertices on sides PQ, QR, RS

                                                   + Number of triangle with vertices on sides QR, SR, SP
                                                   + Number of triangle with vertices on sides RS, PS, PQ
                                                   + Number of triangle with vertices on sides PS, PQ, QR
\[\therefore \] Number of triangle= Number of triangle with vertices on sides
                                         (PQ, QR, RS + QR, SR, SP + RS, PS, PQ + PS, PQ, QR)
Number of triangle\[={}^{3}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{5}{{C}_{1}}+{}^{3}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{5}{{C}_{1}}+{}^{5}{{C}_{1}}\times {}^{6}{{C}_{1}}\times {}^{3}{{C}_{1}}+{}^{6}{{C}_{1}}\times {}^{3}{{C}_{1}}\times {}^{4}{{C}_{1}}\]

They are of the form \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]

\[\begin{align}
  & {}^{3}{{C}_{1}}=\dfrac{3!}{\left( 3-1 \right)!1!}=\dfrac{3!}{2!};{}^{5}{{C}_{1}}=\dfrac{5!}{\left( 5-1 \right)!1!}=\dfrac{5!}{4!} \\
 & {}^{4}{{C}_{1}}=\dfrac{4!}{\left( 4-1 \right)!1!}=\dfrac{4!}{3!};{}^{6}{{C}_{1}}=\dfrac{6!}{\left( 6-1 \right)!1!}=\dfrac{6!}{5!} \\
\end{align}\]

\[\therefore \] Total number of triangles \[=\left( \dfrac{3!}{2!}\times \dfrac{4!}{3!}\times \dfrac{5!}{4!} \right)+\left( \dfrac{4!}{3!}\times \dfrac{5!}{4!}\times \dfrac{6!}{5!} \right)+\left( \dfrac{5!}{4!}\times \dfrac{6!}{5!}\times \dfrac{3!}{2!} \right)+\left( \dfrac{6!}{5!}\times \dfrac{3!}{2!}\times \dfrac{4!}{3!} \right)\]

Cancel out like terms and simplify it
\[=\dfrac{5!}{2!}+\dfrac{6!}{3!}+\dfrac{6!3!}{4!2!}+\dfrac{6!4!}{5!2!}\]

\[\begin{align}
  & \because 1!=1 \\
 & 2!=2\times 1 \\
 & 3!=3\times 2\times 1 \\
\end{align}\]

\[\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & 5!=5\times 4\times 3\times 2\times 1 \\
 & 6!=6\times 5\times 4\times 3\times 2\times 1 \\
 & =\left( \dfrac{5\times 4\times 3\times 2\times 1}{2\times 1} \right)+\left( \dfrac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1} \right)+\left( \dfrac{6\times 5\times 4!\times 3\times 2!}{4!2!} \right)+\left( \dfrac{6\times 5!\times 4\times 3\times 2!}{5!2!} \right) \\
 & =\left( 5\times 4\times 3 \right)+\left( 6\times 5\times 4 \right)+\left( 6\times 5\times 3 \right)+\left( 6\times 4\times 3 \right) \\
 & =60+120+90+72=342 \\
\end{align}\]
\[\therefore \] Total number of triangles = 342
Hence, the correct option is (d) 342.

Note: Take the 3 sides of the convex quadrilateral each time to bring out the number of triangles at those vertices.