The number of elements in the set S = { (r, k) : k is an integer and 36C(r+1) = 6 * (35Cr) / (k^2-3) }, is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q7 – JEE-MAIN 2026

← Back

Home Maths Permutations and CombinationsQuestion #314

The number of elements in the set

S = {(r, k) : k \in Z {and}^{36} C_{r + 1} = \frac{6 (^{35} C_{r})}{(k^{2} - 3)}}

, is:

Clue 1 of 3: Binomial Coefficient Identity

Recall the identity relating binomial coefficients: $^{n} C_{r} = {\frac{n}{r}}^{n - 1} C_{r - 1}$ .

🧠 Full Solution Path

Step 1: Simplify the given equation using binomial identity✦ Active

The given equation is $^{36} C_{r + 1} = \frac{6 (^{35} C_{r})}{(k^{2} - 3)}$ . We use the identity $^{n} C_{r} = {\frac{n}{r}}^{n - 1} C_{r - 1}$ . Applying this to $^{36} C_{r + 1}$ :

^{36} C_{r + 1} = {\frac{36}{r + 1}}^{35} C_{r}

Substitute this into the original equation:

{\frac{36}{r + 1}}^{35} C_{r} = \frac{6 (^{35} C_{r})}{(k^{2} - 3)}

Assuming $^{35} C_{r} \neq 0$ (which implies $0 \leq r \leq 35$ ), we can cancel it from both sides:

\frac{36}{r + 1} = \frac{6}{k^{2} - 3}

Simplify further:

\frac{6}{r + 1} = \frac{1}{k^{2} - 3}

Cross-multiply to express $r$ in terms of $k$ :

6 (k^{2} - 3) = r + 1 ⟹ r = 6 k^{2} - 18 - 1 ⟹ r = 6 k^{2} - 19

💡 Teacher's Secret Hint

Remember to consider the domain of $r$ for binomial coefficients, which is $0 \leq r \leq n$ .

Step 2: Determine the valid range for k○ Expand

For $^{35} C_{r}$ to be defined, $r$ must satisfy $0 \leq r \leq 35$ . Substitute the expression for $r$ into this inequality:

0 \leq 6 k^{2} - 19 \leq 35

This gives two inequalities:

1) $0 \leq 6 k^{2} - 19 ⟹ 19 \leq 6 k^{2} ⟹ k^{2} \geq \frac{19}{6} \approx 3.16$

2) $6 k^{2} - 19 \leq 35 ⟹ 6 k^{2} \leq 54 ⟹ k^{2} \leq 9$

Combining these, we get $\frac{19}{6} \leq k^{2} \leq 9$ . Since $k \in Z$ , $k^{2}$ must be a perfect square. The integer values for $k^{2}$ that satisfy this range are $k^{2} = 4$ and $k^{2} = 9$ .

Also, the denominator $k^{2} - 3$ cannot be zero. $k^{2} - 3 = 0 ⟹ k^{2} = 3$ , which is not possible for integer $k$ . So, this condition is satisfied.

💡 Teacher's Secret Hint

Ensure $k^{2} - 3 \neq 0$ for the expression to be defined. Also, remember $k$ is an integer.

Step 3: Find the pairs (r, k) and count them○ Expand

Case 1: $k^{2} = 4$

$k = 2$ or $k = - 2$ . For both values, $r = 6 (4) - 19 = 24 - 19 = 5$ . This gives two pairs: $(5, 2)$ and $(5, - 2)$ .

Case 2: $k^{2} = 9$

$k = 3$ or $k = - 3$ . For both values, $r = 6 (9) - 19 = 54 - 19 = 35$ . This gives two pairs: $(35, 3)$ and $(35, - 3)$ .

The set $S$ contains the elements: ${(5, 2), (5, - 2), (35, 3), (35, - 3)}$ . The number of elements in the set $S$ is 4.

💡 Teacher's Secret Hint

Each distinct pair $(r, k)$ contributes one element to the set $S$ . Be careful not to miss negative integer values for $k$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 7 – JEE-MAIN 2026

Clue 1 of 3: Binomial Coefficient Identity

🧠 Full Solution Path