Problem Understanding

We are given a triangle ABC and a set of points on its sides: 4 points on side AB, 5 points on side BC, and 4 points on side AC. A total of $4 + 5 + 4 = 13$ points are available. We need to find the total number of distinct pentagons (polygons with 5 vertices) that can be formed by choosing vertices exclusively from these 13 points.

Condition for Forming a Pentagon

To form a non-degenerate pentagon, the 5 chosen vertices cannot have 3 or more points lying on the same straight line. Since the points are located on the sides of the triangle ABC, the only possible collinear sets of 3 or more points are those that lie entirely on one side (AB, BC, or AC).

Therefore, a valid pentagon can only be formed if, out of the 5 chosen vertices, at most 2 vertices come from any single side of the triangle ABC.

Let $n_{AB}$, $n_{BC}$, and $n_{AC}$ be the number of vertices chosen from sides AB, BC, and AC, respectively. For a valid pentagon:

• $n_{AB} + n_{BC} + n_{AC} = 5$ (Total vertices must be 5)
• $n_{AB} \le 2$
• $n_{BC} \le 2$
• $n_{AC} \le 2$

These conditions also respect the maximum number of points available on each side ($N_{AB}=4, N_{BC}=5, N_{AC}=4$).

Calculating Valid Vertex Distributions

We need to find combinations of $(n_{AB}, n_{BC}, n_{AC})$ that satisfy the conditions above. The only possible combinations summing to 5, with each value less than or equal to 2, are permutations of (2, 2, 1).

1. Case 1: (2 points from AB, 2 points from BC, 1 point from AC) i.e., $(n_{AB}=2, n_{BC}=2, n_{AC}=1)$
2. Case 2: (2 points from AB, 1 point from BC, 2 points from AC) i.e., $(n_{AB}=2, n_{BC}=1, n_{AC}=2)$
3. Case 3: (1 point from AB, 2 points from BC, 2 points from AC) i.e., $(n_{AB}=1, n_{BC}=2, n_{AC}=2)$

Calculating Number of Pentagons for Each Case

We use combinations, denoted as $C(n, k)$ or $\binom{n}{k}$, calculated as $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

1. Case 1: (2, 2, 1)
   Number of ways = (Ways to choose 2 from AB) $\times$ (Ways to choose 2 from BC) $\times$ (Ways to choose 1 from AC)
   Number of ways = $\binom{4}{2} \times \binom{5}{2} \times \binom{4}{1}$
   Number of ways = $6 \times 10 \times 4 = 240$.
2. Case 2: (2, 1, 2)
   Number of ways = (Ways to choose 2 from AB) $\times$ (Ways to choose 1 from BC) $\times$ (Ways to choose 2 from AC)
   Number of ways = $\binom{4}{2} \times \binom{5}{1} \times \binom{4}{2}$
   Number of ways = $6 \times 5 \times 6 = 180$.
3. Case 3: (1, 2, 2)
   Number of ways = (Ways to choose 1 from AB) $\times$ (Ways to choose 2 from BC) $\times$ (Ways to choose 2 from AC)
   Number of ways = $\binom{4}{1} \times \binom{5}{2} \times \binom{4}{2}$
   Number of ways = $4 \times 10 \times 6 = 240$.

Total Number of Pentagons

The total number of possible pentagons is the sum of the ways calculated for each valid case:

Total Pentagons = Case 1 + Case 2 + Case 3

Total Pentagons = $240 + 180 + 240 = 660$.