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How many consecutive factors remain in the expansion of ( \frac{(n+5)!}{(n+1)!} )?( \frac{(n+5)!}{(n+1)!} ) के विस्तार में कितने लगातार गुणक बचते हैं?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (4)(4)
The expansion is ((n+5)(n+4)(n+3)(n+2)), so (4) factors remain. The factorial gap gives the number of factors.विस्तार ((n+5)(n+4)(n+3)(n+2)) है, इसलिए (4) गुणक बचते हैं। फैक्टोरियल अंतर से गुणकों की संख्या मिलती है।
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