यदि ( \frac{(n!)2}{(n-3)!(n+3)!}=\frac{5}{21} ), तो (n) का मान क्या है?

If ( \frac{(n!)2}{(n-3)!(n+3)!}=\frac{5}{21} ), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

B. (6)

Step 1

Concept

Putting (n=6) in the simplified form gives \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \). Cancel while checking options.

Step 2

Why this answer is correct

The correct answer is B. (6). Putting (n=6) in the simplified form gives \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \). Cancel while checking options.

Step 3

Exam Tip

सरल रूप में (n=6) रखने पर \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \) मिलता है। विकल्प जांचते समय कटौती करें।

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Mathematics Answer, Explanation and Revision Hints

यदि ( \frac{(n!)2}{(n-3)!(n+3)!}=\frac{5}{21} ), तो (n) का मान क्या है? / If ( \frac{(n!)2}{(n-3)!(n+3)!}=\frac{5}{21} ), what is the value of (n)?

Correct Answer: B. (6). Explanation: सरल रूप में (n=6) रखने पर \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \) मिलता है। विकल्प जांचते समय कटौती करें। / Putting (n=6) in the simplified form gives \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \). Cancel while checking options.

Which concept should I revise for this Mathematics MCQ?

Putting (n=6) in the simplified form gives \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \). Cancel while checking options.

What exam hint can help solve this Mathematics question?

सरल रूप में (n=6) रखने पर \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \) मिलता है। विकल्प जांचते समय कटौती करें।