Given that $ABCD$ is a convex quadrilateral. $3,4,5$ and $6$ points are marked on the sides $AB,BC,CD$ and $DA$ respectively. The no. of triangles with vertices on different sides are?
A. $270$
B. $220$
C. $282$
D. $342$

Hint: In this question, for determining the number of triangles with vertices on different sides, we have to consider four different possibilities of vertices. For this, we have to use the concept of combination. After that, we can find the total of all the possibilities.

Complete step by step solution:
We know that there $3$ points on$AB$, $4$ points on $BC$, $5$ points on $CD$, and $6$ points on $AC$.
Here, we need to consider three points from different sides of a convex quadrilateral that consists of $3,4,5$and $6$points on sides $AB,BC,CD$and $DA$respectively.
So, consider the following four possibilities of vertices using the concept of combination.
1) $AB,BC,CD$
Thus, we get
 $
  {}^3{C_1} \times {}^4{C_1} \times {}^5{C_1} \\
   \Rightarrow \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \times \dfrac{{5!}}{{1!\left( {5 - 1} \right)!}} \\
   \Rightarrow \dfrac{{3!}}{{1!\left( 2 \right)!}} \times \dfrac{{4!}}{{1!\left( 3 \right)!}} \times \dfrac{{5!}}{{1!\left( 4 \right)!}} \\
   \Rightarrow 3 \times 4 \times 5 \\
   \Rightarrow 60 \\
 $
2) $BC,CD,DA$
Thus, we get
 $
  {}^4{C_1} \times {}^5{C_1} \times {}^6{C_1} \\
   \Rightarrow \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \times \dfrac{{5!}}{{1!\left( {5 - 1} \right)!}} \times \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}} \\
   \Rightarrow \dfrac{{4!}}{{1!\left( 3 \right)!}} \times \dfrac{{5!}}{{1!\left( 4 \right)!}} \times \dfrac{{6!}}{{1!\left( 5 \right)!}} \\
   \Rightarrow 4 \times 5 \times 6 \\
   \Rightarrow 120 \\
 $
3) $CD,DA,AB$
Thus, we get
$
  {}^5{C_1} \times {}^6{C_1} \times {}^3{C_1} \\
   \Rightarrow \dfrac{{5!}}{{1!\left( {5 - 1} \right)!}} \times \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}} \times \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \\
   \Rightarrow \dfrac{{5!}}{{1!\left( 4 \right)!}} \times \dfrac{{6!}}{{1!\left( 5 \right)!}} \times \dfrac{{3!}}{{1!\left( 2 \right)!}} \\
   \Rightarrow 5 \times 6 \times 3 \\
   \Rightarrow 90 \\
 $
4) $DA,AB,BC$
Thus, we get
 $
  {}^6{C_1} \times {}^3{C_1} \times {}^4{C_1} \\
   \Rightarrow \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}} \times \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \\
   \Rightarrow \dfrac{{6!}}{{1!\left( 5 \right)!}} \times \dfrac{{3!}}{{1!\left( 2 \right)!}} \times \dfrac{{4!}}{{1!\left( 3 \right)!}} \\
   \Rightarrow 6 \times 3 \times 4 \\
   \Rightarrow 72 \\
 $
Now, we will add these values to find the number of triangles.
Thus, we get
${\text{Number of triangles }} = {\text{ }}60 + 120 + 90 + 72 = 342$
Thus, there are $342$ triangles with vertices on different sides.

Option ‘D’ is correct

Additional Information: Combinations are a mathematical technique of picking items or numbers from a set or group of things in such a manner that the sequence of the objects is irrelevant.

Note: Many students make mistakes in understanding the term convex quadrilateral. A convex quadrilateral is a four-sided polygon with inner angles of fewer than $180$ degrees. Also, the two diagonals of a convex quadrilateral are fully contained inside the figure.