Question

For a quadrilateral $ABCD$ , if $3,4,5$ and $6$ points are marked on the sides $AB,BC,CD$ and $DA$ respectively. Then number of triangles that can be formed with vertices on different sides, is

• A. $270$
• B. $342$
• C. $282$
• D. $252$

Answer 1

The correct option is B $342$

Required triangle must have one point on each side of Quadrilateral.
Hence the required number of triangles
$={}^3{C}_1\times {}^4{C}_1\times {}^5{C}_1+{}^3{C}_1\times {}^4{C}_1\times {}^6{C}_1+{}^3{C}_1\times {}^5{C}_1\times {}^6{C}_1+{}^4{C}_1\times {}^5{C}_1\times {}^6{C}_1$
$=342$