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Combination

If n-1C3+n-...

Question

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

A

n \geq 4

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B

n > 5

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C

n > 7

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D

none of these.

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Solution

The correct option is A $n > 7$
We have,
$\begin{matrix} ^{n - 1} C_{3} +^{n - 1} C_{4} >^{n} C_{3}^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}^{n} C_{4} >^{n} C_{3} \frac{n!}{4! n - 4!} > \frac{n!}{3! n - 3!} \frac{n (n - 1) (n - 2) (n - 3)}{4 \times 3!} > \frac{n (n - 1) (n - 2)}{3!} n - 3 > 4 H e n c e, n > 7 S o o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

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

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

, then the least value of

n

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

, then set of complete values of n is:

^{(n - 1)} C_{3} +^{(n - 1)} C_{4} >^{n} C_{3}

, then the minimum value of

n

^{n} C_{3} +^{n} C_{4}

^{n + 1} C_{3},

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