Understanding the Triangle Formation Problem

The question asks us to find the total number of triangles that can be formed by selecting vertices from a specific set of points. These points are located on the sides of a larger triangle, named \(PQR\). We are given:

• \(3\) points on the side \(PQ\).
• \(4\) points on the side \(QR\).
• \(5\) points on the side \(RP\).

The key is to determine the total number of points available and identify any sets of points that cannot form a triangle (i.e., sets of collinear points).

Based on the typical interpretation of such problems and aiming to match the provided answer options, we assume that the vertices \(P, Q, R\) of the main triangle \(PQR\) are *not* included in the points used to form the new triangles. The vertices for the new triangles must be chosen *exclusively* from the \(3 + 4 + 5 = 12\) specified points lying on the sides.

Calculating Total Possible Combinations of 3 Points

To form a triangle, we need to choose any 3 points. If all points were non-collinear, the total number of triangles would simply be the number of ways to choose 3 points from the total available points. The total number of points available for selection is \(N = 3 + 4 + 5 = 12\).

The number of ways to choose 3 points from 12 is given by the combination formula: \(\binom{N}{k} = \binom{12}{3}\) Calculating this value:

\(\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220\)

So, there are \(220\) ways to select 3 points from the 12 available points.

Identifying and Subtracting Collinear Combinations

However, a triangle cannot be formed if the 3 selected points lie on the same straight line (are collinear). The points on each side of the triangle \(PQR\) are collinear.

• The \(3\) points on side \(PQ\) are collinear.
• The \(4\) points on side \(QR\) are collinear.
• The \(5\) points on side \(RP\) are collinear.

We need to calculate how many combinations of 3 points can be chosen from each of these sets and subtract them from the total combinations.

• Number of combinations from the 3 points on \(PQ\): \(\binom{3}{3} = \frac{3!}{3!(3-3)!} = 1\)
• Number of combinations from the 4 points on \(QR\): \(\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4\)
• Number of combinations from the 5 points on \(RP\): \(\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10\)

The total number of combinations of 3 collinear points is \(1 + 4 + 10 = 15\).

Final Calculation for the Number of Triangles

To find the number of actual triangles formed, we subtract the number of collinear combinations from the total number of combinations calculated earlier.

Number of triangles = (Total combinations of 3 points) - (Combinations of 3 collinear points)

Number of triangles = \(220 - 15\)

Number of triangles = \(205\)

Therefore, \(205\) triangles can be formed using the specified points as vertices.