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 $p_1, p_2, \ldots , p_{13}$, is ______
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Solution Steps
1
Total ways to choose 5 points
Total points = $4 + 5 + 4 = 13$.
Total ways to choose any 5 points:
$^{13}C_5 = 1287$.
2
When does selection NOT form a pentagon?
A pentagon requires no three selected points to be collinear.
Points on the same side of triangle are collinear.
So invalid cases occur when at least 3 points are chosen from the same side.
3
All 5 from same side
Possible only from side $BC$ (5 points).
Number of ways: $^{5}C_5 = 1$.
4
Exactly 4 from one side
From $AB$: $^{4}C_4 \times ^9C_1 = 1 \times 9 = 9$
From $AC$: $^{4}C_4 \times ^9C_1 = 9$
From $BC$: $^{5}C_4 \times ^8C_1 = 5 \times 8 = 40$
Total = $9 + 9 + 40 = 58$.
5
Exactly 3 from one side
From $AB$: $^{4}C_3 \times ^9C_2 = 4 \times 36 = 144$
From $AC$: same = 144
From $BC$: $^{5}C_3 \times ^8C_2 = 10 \times 28 = 280$
Total = $144 + 144 + 280 = 568$.
6
Total invalid selections
$1 + 58 + 568 = 627$.
7
Valid pentagons
$1287 – 627 = \boxed{660}$.
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Theory Topics
1. Counting Using Combinations
In combinatorics, when we need to select $r$ objects from $n$ distinct objects without regard to order, we use combinations given by $^nC_r = \frac{n!}{r!(n-r)!}$. In geometry counting problems, this helps in determining the number of polygons formed by selecting vertices from a set of points. Here, total 13 points were available, so total selections of 5 points were computed using $^{13}C_5$. This represents all possible subsets of size five. However, not every subset forms a valid pentagon. Thus, combinations provide the starting point, and then invalid geometric configurations are subtracted. Understanding combinations is fundamental for JEE Main problems involving polygon counting, selection problems, and arrangements under constraints.
2. Collinearity and Polygon Formation
A necessary condition for forming an $n$-sided polygon is that no three chosen vertices are collinear. If three or more points lie on a straight line, the resulting figure degenerates and does not form a proper polygon. In this problem, points on each side of the triangle are collinear. Therefore, if we select three or more points from the same side, those three points lie on a straight line and cannot contribute to a pentagon. Recognizing such geometric constraints is essential in counting problems. Instead of directly counting valid cases, it is often easier to count all possibilities and subtract invalid collinear selections.
3. Principle of Complementary Counting
Complementary counting is a powerful strategy where instead of directly counting valid outcomes, we count total outcomes and subtract invalid ones. In this question, directly counting valid pentagons would be complex. Instead, we first counted total ways to choose any five points from thirteen. Then, we counted selections that contain at least three collinear points, since those cannot form pentagons. By subtracting invalid cases from total combinations, we obtained the required count. This technique simplifies complex counting problems and is frequently used in JEE Main combinatorics questions.
4. Systematic Case Analysis
When dealing with restrictions like “at least three from the same side,” systematic case classification avoids double counting. We divided invalid cases into: five from one side, four from one side, and exactly three from one side. Each case was mutually exclusive and collectively exhaustive. Careful multiplication using combinations ensured accuracy. In competitive exams, structured casework prevents errors and ensures no configuration is missed. Mastery of such structured counting techniques is essential for high scores in combinatorics problems in JEE Main.
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FAQs
Q
What is the total number of ways to choose 5 points from 13?
$^{13}C_5 = 1287$.
Q
Why subtract cases with 3 collinear points?
Because three collinear points cannot form a proper pentagon.
Q
Why are points on same side collinear?
They lie on a straight line segment of the triangle.
Q
How many total invalid cases were there?
Total invalid selections were 627.
Q
What is complementary counting?
Counting total outcomes and subtracting unwanted ones.
Q
Why consider exactly 3, 4, and 5 from one side?
Because at least three collinear points invalidate polygon formation.
Q
Final answer?
660.
Q
Is order of vertices important?
No, we only select sets of 5 points.
Q
Which side had 5 points?
Side BC.
Q
Which formula used for combinations?
$^nC_r = \frac{n!}{r!(n-r)!}$.