Question

ABCD is a convex quadrilateral and 3, 4, 5 and 6 are marked on the sides AB, BC, CD and DA respectively. The number of triangles with vertices on different sides is (none of them is A, B, c or D)
A. 270
B. 220
C. 282
D. 342

Answer 1

Hint: Here in this question we have to determine the number of triangles formed by connecting the points which are present on the convex quadrilateral. To find this we use the concept of combination and simple arithmetic operations and hence we obtain the solution for the given question.

Complete step by step solution:
A convex quadrilateral is a four-sided polygon that has an interior angle that measures less than \[{180^ \circ }\] each. By the data of the given question. There is 3 points on the line AB. There is 4 points on the line BC. There is 5 points on the line CD and there is 6 points on the line DA. Here we have to construct the triangle, the triangle is formed only by three points, but here we have 4 points. So, we have to consider the 3 points and construct the triangle.
Let us consider the points which lie on AB, BC and CD. We use the concept of combination.
\[{}^3{C_1} \times {}^4{C_1} \times {}^5{C_1}\]
As we know that the formula of combination \[{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], so we can write it has
\[ \Rightarrow \dfrac{{3!}}{{(3 - 1)!1!}} \times \dfrac{{4!}}{{(4 - 1)!1!}} \times \dfrac{{5!}}{{(5 - 1)!1!}}\]
On simplifying this we can write it as
\[ \Rightarrow \dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{3!}} \times \dfrac{{5!}}{{4!}}\]
The formula for \[n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1\] applying this we write it as
\[ \Rightarrow \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{4 \times 3!}}{{3!}} \times \dfrac{{5 \times 4!}}{{4!}}\]
On simplifying this
\[ \Rightarrow 3 \times 4 \times 5\]
\[ \Rightarrow 60\]
Let we consider the points which lie on BC, CD and DA. We use the concept of combination.
\[{}^4{C_1} \times {}^5{C_1} \times {}^6{C_1}\]
As we know that the formula of combination \[{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], so we can write it has
\[ \Rightarrow \dfrac{{4!}}{{(4 - 1)!1!}} \times \dfrac{{5!}}{{(5 - 1)!1!}} \times \dfrac{{6!}}{{(6 - 1)!1!}}\]
On simplifying this we can write it as
\[ \Rightarrow \dfrac{{4!}}{{3!}} \times \dfrac{{5!}}{{4!}} \times \dfrac{{6!}}{{5!}}\]
The formula for \[n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1\] applying this we write it as
\[ \Rightarrow \dfrac{{4 \times 3!}}{{3!}} \times \dfrac{{5 \times 4!}}{{4!}} \times \dfrac{{6 \times 5!}}{{5!}}\]
On simplifying this
\[ \Rightarrow 4 \times 5 \times 6\]
\[ \Rightarrow 120\]
Let we consider the points which lie on CD, DA and AB. We use the concept of combination.
\[{}^5{C_1} \times {}^6{C_1} \times {}^3{C_1}\]
As we know that the formula of combination \[{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], so we can write it has
\[ \Rightarrow \dfrac{{5!}}{{(5 - 1)!1!}} \times \dfrac{{6!}}{{(6 - 1)!1!}} \times \dfrac{{3!}}{{(3 - 1)!1!}}\]
On simplifying this we can write it as
\[ \Rightarrow \dfrac{{5!}}{{4!}} \times \dfrac{{6!}}{{5!}} \times \dfrac{{3!}}{{2!}}\]
The formula for \[n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1\] applying this we write it as
\[ \Rightarrow \dfrac{{5 \times 4!}}{{4!}} \times \dfrac{{6 \times 5!}}{{5!}} \times \dfrac{{3 \times 2!}}{{2!}}\]
On simplifying this
\[ \Rightarrow 5 \times 6 \times 3\]
\[ \Rightarrow 90\]
Let we consider the points which lie on DA, AB and AB. We use the concept of combination.
\[{}^6{C_1} \times {}^3{C_1} \times {}^4{C_1}\]
As we know that the formula of combination \[{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], so we can write it has
\[ \Rightarrow \dfrac{{6!}}{{(6 - 1)!1!}} \times \dfrac{{3!}}{{(3 - 1)!1!}} \times \dfrac{{4!}}{{(4 - 1)!1!}}\]
On simplifying this we can write it as
\[ \Rightarrow \dfrac{{6!}}{{5!}} \times \dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{3!}}\]
The formula for \[n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1\] applying this we write it as
\[ \Rightarrow \dfrac{{6 \times 5!}}{{5!}} \times \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{4 \times 3!}}{{3!}}\]
On simplifying this
\[ \Rightarrow 6 \times 3 \times 4\]
\[ \Rightarrow 72\]
Now we add all the possibilities and hence determine the solution for the question.
On adding we have
\[60{\text{ }} + {\text{ }}120{\text{ }} + {\text{ }}90{\text{ }} + {\text{ }}72\]
On simplifying we get
\[ \Rightarrow 342\]

So, the correct answer is Option D.

Note: A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. We must know the formula \[{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\] to simplify and we use simple mathematical operations.