If $$C(n-1$$,$$r)=k2-3$$·$$C(n$$,$$r+1)$$ then k∈

Correct option is 4−2,−3∪3,2

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$I f C (n - 1, r) = (k^{2} - 3) \cdot C (n, r + 1) t h e n k \in$

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−∞,−2

2,∞

3,2

−2,−3∪3,2

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detailed solution

Correct option is D

Given Cn−1,r=k2−3Cn,r+1it gives n−1!r!n−r−1!=k2−3n!r+1!n−r−1!k2−3=r+1n0

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If C(n-1,r)=k2-3·C(n,r+1) then k∈

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$I f C (n - 1, r) = (k^{2} - 3) \cdot C (n, r + 1) t h e n k \in$