Question ID: #748
Let ABC be a triangle. Consider four points $p_{1}, p_{2}, p_{3}, p_{4}$ on the side AB, five points $p_{5},p_{6},p_{7},p_{8},p_{9}$ on the side BC and four points $p_{10},p_{11},p_{12},p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_{1},p_{2},…,p_{13}$, is
Solution:
Total points $N = 4 + 5 + 4 = 13$.
Points on side AB ($n_1$) = 4.
Points on side BC ($n_2$) = 5.
Points on side AC ($n_3$) = 4.
To form a pentagon, we need to select 5 vertices such that no three are collinear. Since the points lie on the sides of a triangle, points on the same side are collinear. Thus, we can select at most 2 points from any single side.
We need to partition the number 5 into three parts $(x, y, z)$ such that $x+y+z = 5$ and $0 \le x, y, z \le 2$.
The only possible combination for $(x, y, z)$ is a permutation of $(2, 2, 1)$.
**Case 1:** 2 points from AB, 2 points from BC, 1 point from AC.
$$ \binom{4}{2} \times \binom{5}{2} \times \binom{4}{1} = 6 \times 10 \times 4 = 240 $$
**Case 2:** 2 points from AB, 1 point from BC, 2 points from AC.
$$ \binom{4}{2} \times \binom{5}{1} \times \binom{4}{2} = 6 \times 5 \times 6 = 180 $$
**Case 3:** 1 point from AB, 2 points from BC, 2 points from AC.
$$ \binom{4}{1} \times \binom{5}{2} \times \binom{4}{2} = 4 \times 10 \times 6 = 240 $$
Total number of pentagons = $240 + 180 + 240 = 660$.
Ans. (660)
Was this solution helpful?
YesNo