The least value of n such that { }^{(n-1)} C_ | vedclass

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(C) Given the inequality: ${ }^{(n-1)} C_3 + { }^{(n-1)} C_4 > { }^n C_3$ Using the Pascal's identity ${ }^n C_{r-1} + { }^n C_r = { }^{n+1} C_r$,we h

Vedclass · Accepted Answer

$8$

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(C) Given the inequality: ${ }^{(n-1)} C_3 + { }^{(n-1)} C_4 > { }^n C_3$ Using the Pascal's identity ${ }^n C_{r-1} + { }^n C_r = { }^{n+1} C_r$,we have: ${ }^{(n-1)} C_3 + { }^{(n-1)} C_4 = { }^n C_4$ Substituting this into the inequality: ${ }^n C_4 > { }^n C_3$ Expanding the combinations: $\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$ Since $n! > 0$,we can divide both sides by $n!$: $\frac{1}{4!(n-4)!} > \frac{1}{3!(n-3)!}$ $\frac{1}{4 	imes 3! 	imes (n-4)!} > \frac{1}{3! 	imes (n-3) 	imes (n-4)!}$ $\frac{1}{4} > \frac{1}{n-3}$ Since $n-3 > 0$ (as $n \ge 4$),we can cross-multiply: $n - 3 > 4$ $n > 7$ Therefore,the least integer value of $n$ is $8$.

The least value of $n$ such that ${ }^{(n-1)} C_3 + { }^{(n-1)} C_4 > { }^n C_3$ is:

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