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Question

The number of positive integers satisfying the inequality
n+1Cn2n+1Cn1100 is

A
One
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B
Eight
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C
Five
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D
None of these
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Solution

The correct option is B Eight
Given, (n+1)!3!(n2)!(n+1)!2!.(n1)!
=(n1)(n)(n+1)6n(n+1)2
=(n)(n+1)n136
(n)(n+1)n46100
n(n+1)(n4)6100
Since nϵN
Hence
2,3,4,5,6,7,8,9 satisfy the above inequality.

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