Maths Q8 – JEE-MAIN 2025

Home Maths Permutations and CombinationsQuestion #331

Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If

p + q = 126

, then the eccentricity of the ellipse

\frac{x^{2}}{16} + \frac{y^{2}}{n} = 1

is:

Clue 1 of 3: Combinations for Polygons

The number of ways to form a polygon with $k$ vertices from $n$ available vertices is given by the combination formula $^{n} C_{k}$ .

🧠 Full Solution Path

Step 1: Express p and q in terms of n✦ Active

The number of triangles (p) formed from $n$ vertices is $^{n} C_{3}$ . The number of quadrilaterals (q) formed from $n$ vertices is $^{n} C_{4}$ .

p = (\binom{n}{3}) = \frac{n (n - 1) (n - 2)}{6}

and q = (\binom{n}{4}) = \frac{n (n - 1) (n - 2) (n - 3)}{24}

Step 2: Solve for n using the given condition○ Expand

Given $p + q = 126$ . Using the combinatorial identity $(\binom{n}{r}) + (\binom{n}{r + 1}) = (\binom{n + 1}{r + 1})$ , we can simplify $p + q$ .

(\binom{n}{3}) + (\binom{n}{4}) = (\binom{n + 1}{4})

So, (\binom{n + 1}{4}) = 126

By inspection or calculation, we find that (\binom{9}{4}) = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126.

Therefore, n + 1 = 9 ⟹ n = 8.

Step 3: Calculate the eccentricity of the ellipse○ Expand

Substitute $n = 8$ into the ellipse equation $\frac{x^{2}}{16} + \frac{y^{2}}{n} = 1$ . The equation becomes $\frac{x^{2}}{16} + \frac{y^{2}}{8} = 1$ . For an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where $a^{2} > b^{2}$ , the eccentricity $e$ is given by $b^{2} = a^{2} (1 - e^{2})$ .

a^{2} = 16, b^{2} = 8

8 = 16 (1 - e^{2})

\frac{1}{2} = 1 - e^{2}

e^{2} = 1 - \frac{1}{2} = \frac{1}{2}

e = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}

💡 Teacher's Secret Hint

Ensure to correctly identify $a^{2}$ and $b^{2}$ based on which term is larger, as this determines the orientation of the major axis.

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Maths Question 8 – JEE-MAIN 2025

Clue 1 of 3: Combinations for Polygons

🧠 Full Solution Path