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How many consecutive factors remain in \(\frac{(n+3)!}{n!}\)?\(\frac{(n+3)!}{n!}\) में कुल कितने क्रमागत गुणक बचते हैं?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(3\)\(3\)
Expanding \((n+3)!\) up to \(n!\) gives \((n+3)!=(n+3)(n+2)(n+1)n!\). After cancelling \(n!\), we get \(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)\), which contains 3 consecutive factors. There would be 4 factors in \(\frac{(n+4)!}{n!}\). Exam tip: In \(\frac{(n+r)!}{n!}\), exactly \(r\) consecutive factors remain.\((n+3)!\) को \(n!\) तक फैलाने पर \((n+3)!=(n+3)(n+2)(n+1)n!\) प्राप्त होता है। इसलिए \(n!\) कटने पर \(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)\) बचता है, जिसमें 3 क्रमागत गुणक हैं। \(4\) गुणक \(\frac{(n+4)!}{n!}\) में होते। परीक्षा टिप: \(\frac{(n+r)!}{n!}\) में सामान्यतः \(r\) क्रमागत गुणक बचते हैं।
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