1. Key Concepts and Formulas

This problem is a classic example of how different branches of mathematics, specifically Combinatorics and Conic Sections (Ellipses), are integrated in JEE. To solve it efficiently and accurately, we need a solid understanding of the following concepts:

• Combinations ($^n C_k$): This formula is used to calculate the number of ways to choose $k$ distinct items from a set of $n$ distinct items, where the order of selection does not matter. The formula is: $^n C_k = \frac{n!}{k!(n-k)!}$ In geometry, when forming polygons by selecting vertices, the order of vertices does not change the polygon, hence combinations are appropriate. For example, choosing vertices A, B, C forms the same triangle as B, C, A.

• Pascal's Identity: This is a fundamental identity in combinatorics that simplifies sums of consecutive combination terms: $^n C_r + ^n C_{r+1} = ^{n+1} C_{r+1}$ This identity is extremely useful for reducing expressions involving sums of combinations to a single term, often simplifying algebraic manipulation.

• Standard Form of an Ellipse and Eccentricity: An ellipse centered at the origin typically has one of two standard forms:

  1. $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (Major axis along the x-axis)
  2. $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (Major axis along the y-axis) In both cases, $a$ represents the length of the semi-major axis, and $b$ represents the length of the semi-minor axis. Crucially, $a^2$ is always the larger of the two denominators under $x^2$ and $y^2$, and $b^2$ is the smaller one. The eccentricity ($e$) of an ellipse, which measures its "ovalness" or how much it deviates from a perfect circle, is given by the formula: $e = \sqrt{1 - \frac{b^2}{a^2}}$ (Note: $0 < e < 1$ for an ellipse).

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2. Step-by-Step Solution

Step 1: Express the number of triangles ($p$) and quadrilaterals ($q$) using combinations.

• Why this step? The problem asks us to form triangles and quadrilaterals by joining the vertices of a regular polygon of $n$ sides. To form any polygon, we simply need to select a specific number of distinct vertices. The order in which we select these vertices does not change the polygon formed (e.g., selecting vertices V1, V2, V3 creates the same triangle as V2, V1, V3). Therefore, this is a problem of combinations.
• Calculation:
  • To form a triangle, we need to choose 3 distinct vertices out of the $n$ available vertices. $p = ^n C_3$
  • To form a quadrilateral, we need to choose 4 distinct vertices out of the $n$ available vertices. $q = ^n C_4$

Step 2: Use the given condition $p+q=126$ to find the value of $n$.

• Why this step? The problem provides a direct relationship between $p$ and $q$. By substituting our combinatorial expressions for $p$ and $q$ into this equation, we can form an equation solely in terms of $n$. To simplify this equation and solve for $n$ efficiently, Pascal's Identity will be extremely useful.
• Calculation: Substitute the expressions for $p$ and $q$ into the given condition: $^n C_3 + ^n C_4 = 126$ Now, apply Pascal's Identity, which states $^n C_r + ^n C_{r+1} = ^{n+1} C_{r+1}$. Here, we have $r=3$. $^{n+1} C_4 = 126$ To find $n$, we need to solve this combination equation. Let $N = n+1$ for simplicity. So we need to solve $^N C_4 = 126$. Expand the combination formula for $^N C_4$: $\frac{N!}{4!(N-4)!} = 126$ $\frac{N(N-1)(N-2)(N-3)}{4 \times 3 \times 2 \times 1} = 126$ $\frac{N(N-1)(N-2)(N-3)}{24} = 126$ Multiply both sides by 24: $N(N-1)(N-2)(N-3) = 126 \times 24$ $N(N-1)(N-2)(N-3) = 3024$ We are looking for four consecutive integers whose product is 3024. We can estimate or use trial and error:
  • If $N=7$, $7 \times 6 \times 5 \times 4 = 840$ (Too small)
  • If $N=8$, $8 \times 7 \times 6 \times 5 = 1680$ (Still too small)
  • If $N=9$, $9 \times 8 \times 7 \times 6 = 72 \times 42 = 3024$ (This matches!) So, $N=9$. Since we defined $N = n+1$, we have: $n+1 = 9 \implies n = 8$

Step 3: Substitute the value of $n$ into the equation of the ellipse.

• Why this step? The final part of the problem requires us to find the eccentricity of an ellipse whose equation depends on $n$. Now that we have determined $n=8$, we can form the specific equation of the ellipse, which is essential for calculating its eccentricity.
• Calculation: The given ellipse equation is $\frac{x^2}{16} + \frac{y^2}{n} = 1$. Substitute $n=8$: $\frac{x^2}{16} + \frac{y^2}{8} = 1$

Step 4: Identify $a^2$ and $b^2$ for the ellipse.

• Why this step? To calculate the eccentricity using the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$, we must correctly identify the values of $a^2$ (square of the semi-major axis) and $b^2$ (square of the semi-minor axis). It's crucial to remember that $a^2$ is always the larger of the two denominators in the standard ellipse equation, and $b^2$ is the smaller one.
• Calculation: Comparing our ellipse equation $\frac{x^2}{16} + \frac{y^2}{8} = 1$ with the general standard form $\frac{x^2}{D_1} + \frac{y^2}{D_2} = 1$, we have $D_1 = 16$ and $D_2 = 8$. Since $16 > 8$, the larger denominator is $16$ and the smaller is $8$. Therefore: $a^2 = 16$ $b^2 = 8$ (This also tells us that the major axis of this ellipse lies along the x-axis, as $a^2$ is under $x^2$).

Step 5: Calculate the eccentricity ($e$) of the ellipse.

• Why this step? With the correct values for $a^2$ and $b^2$ established, we can now directly apply the eccentricity formula to find the required value.
• Calculation: Using the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$: Substitute $a^2=16$ and $b^2=8$: $e = \sqrt{1 - \frac{8}{16}}$ Simplify the fraction inside the square root: $e = \sqrt{1 - \frac{1}{2}}$ $e = \sqrt{\frac{1}{2}}$ Rationalize the denominator to get the standard form: $e = \frac{1}{\sqrt{2}}$

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3. Important Tips and Common Pitfalls

• Combinations vs. Permutations: Always critically assess whether the order of selection matters. When forming geometric shapes like polygons by joining vertices, the order does not matter, so use combinations ($^n C_k$). Using permutations ($^n P_k$) by mistake is a common error.
• Master Pascal's Identity: The identity $^n C_r + ^n C_{r+1} = ^{n+1} C_{r+1}$ is a powerful tool for simplifying combinatorial expressions. Recognizing when to apply it can save significant time and prevent complex calculations involving factorials.
• Solving for $n$ in Factorial Equations: When faced with an equation like $N(N-1)(N-2)(N-3) = 3024$, trying to solve it as a quartic polynomial is often inefficient and prone to errors. Instead, use estimation (e.g., $N^4 \approx 3024 \implies N \approx \sqrt[4]{3024} \approx 7.4$) and trial-and-error with small integer values.
• Identifying $a^2$ and $b^2$ for Ellipse Eccentricity: This is a crucial step. Never assume that $a^2$ is always the denominator under $x^2$ and $b^2$ is under $y^2$. The definition is that $a^2$ is always the larger of the two denominators, and $b^2$ is the smaller. This distinction determines the orientation of the major axis and is essential for the correct calculation of eccentricity. Failing to identify them correctly is a very common mistake.
• Eccentricity Range: Remember that for an ellipse, the eccentricity $e$ must always satisfy $0 < e < 1$. If your calculated value falls outside this range, it's a strong indicator of an error in your calculations or identification of $a^2$ and $b^2$.

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4. Summary / Key Takeaway

This problem beautifully demonstrates the interconnectedness of different mathematical topics in JEE. The solution required:

1. Applying combinatorics principles to count geometric configurations ($p$ and $q$).
2. Utilizing Pascal's Identity to simplify a combinatorial equation and solve for an unknown variable ($n$).
3. Substituting the derived value into a conic section equation.
4. Correctly identifying the semi-axes of an ellipse and applying the eccentricity formula.

The ability to seamlessly transition between these concepts and apply the correct formulas and identities is key to solving such multi-concept problems.