( \frac{(n+2)!}{(n-4)!} ) के विस्तार में कितने लगातार गुणक होंगे?
How many consecutive factors will be in the expansion of ( \frac{(n+2)!}{(n-4)!} )?
Explanation opens after your attempt
Correct Answer
C. (6)
Concept
The expansion is ((n+2)(n+1)n(n-1)(n-2)(n-3)). Therefore there are (6) consecutive factors.
Why this answer is correct
The correct answer is C. (6). The expansion is ((n+2)(n+1)n(n-1)(n-2)(n-3)). Therefore there are (6) consecutive factors.
Exam Tip
विस्तार ((n+2)(n+1)n(n-1)(n-2)(n-3)) है। इसलिए (6) लगातार गुणक मिलते हैं।
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