यदि ( \frac{n!}{(n-2)!}=9n ), तो (n) का धनात्मक मान क्या है?

If ( \frac{n!}{(n-2)!}=9n ), what is the positive value of (n)?

Explanation opens after your attempt
Correct Answer

B. (10)

Step 1

Concept

The left side is (n(n-1)), so (n(n-1)=9n). Since (n>0), (n-1=9) and (n=10).

Step 2

Why this answer is correct

The correct answer is B. (10). The left side is (n(n-1)), so (n(n-1)=9n). Since (n>0), (n-1=9) and (n=10).

Step 3

Exam Tip

बायां पक्ष (n(n-1)) है, इसलिए (n(n-1)=9n)। (n>0) होने से (n-1=9) और (n=10)।

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Mathematics Answer, Explanation and Revision Hints

यदि ( \frac{n!}{(n-2)!}=9n ), तो (n) का धनात्मक मान क्या है? / If ( \frac{n!}{(n-2)!}=9n ), what is the positive value of (n)?

Correct Answer: B. (10). Explanation: बायां पक्ष (n(n-1)) है, इसलिए (n(n-1)=9n)। (n>0) होने से (n-1=9) और (n=10)। / The left side is (n(n-1)), so (n(n-1)=9n). Since (n>0), (n-1=9) and (n=10).

Which concept should I revise for this Mathematics MCQ?

The left side is (n(n-1)), so (n(n-1)=9n). Since (n>0), (n-1=9) and (n=10).

What exam hint can help solve this Mathematics question?

बायां पक्ष (n(n-1)) है, इसलिए (n(n-1)=9n)। (n>0) होने से (n-1=9) और (n=10)।