If $^{n} C_{3} +^{n} C_{4} >^{n + 1} C_{3}$ , then

n > 6

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n > 7

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n < 6

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none of these

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Solution

The correct option is A $n > 6$
$^{n} C_{3} +^{n} C_{4}$
$=^{n + 1} C_{4}$
$=^{n + 1} C_{n - 3}$
Now
$^{n + 1} C_{n - 3} >^{n + 1} C_{3}$
Hence
$n - 3 > 3$
$n > 6$ .

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