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The least value of n so that $^{n - 1} C_{3} +^{n - 1} C_{4} >^{n} C_{3}$

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answer is C.

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Detailed Solution

$^{n - 1} C_{3} +^{n - 1} C_{4} >^{n} C_{3}$

$^{n} C_{4} >^{n} C_{3}$

$\frac{n!}{(n - 4)! 4!} > \frac{n!}{(n - 3)! 3!}$

$\frac{1}{(n - 4)! 4. 3!} > \frac{1}{(n - 3) . (n - 4)! 3!}$

$\frac{1}{4} > \frac{1}{n - 3}$

$n - 3 > 4$

$n > 7$

Least value of n is 8

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