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If \( \frac{(n+2)!}{n!}=132 \), what is the value of (n)?यदि \( \frac{(n+2)!}{n!}=132 \), तो (n) का मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (10)(10)
It gives \((n+2)(n+1)=132\), and \(12\cdot11=132\), so \(n=10\). Recognize consecutive factors.यह \((n+2)(n+1)=132\) देता है और \(12\cdot11=132\), इसलिए \(n=10\) है। लगातार गुणकों को पहचानें।
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