AP EAMCET · Maths · Permutation Combination
\({ }^{2 \mathrm{n}} \mathrm{C}_4:{ }^{\mathrm{n}} \mathrm{C}_3=99: 4 \Rightarrow \mathrm{n}=\)
- A \(7\)
- B \(6\)
- C \(8\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
\({ }^{2 n} C_4:{ }^n C_3=99: 4\) \(\Rightarrow \frac{\frac{2 n !}{4 !(2 n-4) !}}{\frac{n !}{3 !(n-3) !}}=\frac{99}{4}\) \(\Rightarrow \frac{2 \mathrm{n}(2 \mathrm{n}-1)(2 \mathrm{n}-2)(2 \mathrm{n}-3)}{4 \cdot \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}=\frac{99}{4}\)…
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