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How many consecutive factors remain in \(\frac{(n+5)!}{(n+2)!}\)?\(\frac{(n+5)!}{(n+2)!}\) में कितने क्रमागत गुणक बचते हैं?
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Correct answer:सही उत्तर: 33
Since \((n+5)!=(n+5)(n+4)(n+3)(n+2)!\), cancelling \((n+2)!\) leaves \((n+5)(n+4)(n+3)\). Hence, 3 consecutive factors remain. Four factors would remain if the denominator were \((n+1)!\). Exam tip: in \(\frac{(n+r)!}{n!}\), usually \(r\) factors remain after cancellation.क्योंकि \((n+5)!=(n+5)(n+4)(n+3)(n+2)!\), इसलिए \((n+2)!\) कटने पर \((n+5)(n+4)(n+3)\) बचता है। अतः 3 क्रमागत गुणक बचते हैं। 4 गुणक तभी बचते जब हर में \((n+1)!\) होता। परीक्षा टिप: \(\frac{(n+r)!}{n!}\) में सामान्यतः \(r\) गुणक बचते हैं।
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