यदि ( \frac{(n+5)!}{(n+3)!}=132 ), तो (n) का मान क्या है?

If ( \frac{(n+5)!}{(n+3)!}=132 ), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

B. (7)

Step 1

Concept

((n+5)(n+4)=132) and \(12\cdot11=132\), so (n=7). Find (n) from consecutive factors.

Step 2

Why this answer is correct

The correct answer is B. (7). ((n+5)(n+4)=132) and \(12\cdot11=132\), so (n=7). Find (n) from consecutive factors.

Step 3

Exam Tip

((n+5)(n+4)=132) और \(12\cdot11=132\), इसलिए (n=7) है। लगातार गुणकों से (n) निकालें।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि ( \frac{(n+5)!}{(n+3)!}=132 ), तो (n) का मान क्या है? / If ( \frac{(n+5)!}{(n+3)!}=132 ), what is the value of (n)?

Correct Answer: B. (7). Explanation: ((n+5)(n+4)=132) और \(12\cdot11=132\), इसलिए (n=7) है। लगातार गुणकों से (n) निकालें। / ((n+5)(n+4)=132) and \(12\cdot11=132\), so (n=7). Find (n) from consecutive factors.

Which concept should I revise for this Mathematics MCQ?

((n+5)(n+4)=132) and \(12\cdot11=132\), so (n=7). Find (n) from consecutive factors.

What exam hint can help solve this Mathematics question?

((n+5)(n+4)=132) और \(12\cdot11=132\), इसलिए (n=7) है। लगातार गुणकों से (n) निकालें।