The least value of n for which { }^{(n-1)} C_ | vedclass

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(D) Using the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$,we have ${ }^{(n-1)} C_2 + { }^{(n-1)} C_3 = { }^n C_3$. Given the inequality ${ }^

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$6$

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(D) Using the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$,we have ${ }^{(n-1)} C_2 + { }^{(n-1)} C_3 = { }^n C_3$. Given the inequality ${ }^n C_3 > { }^n C_2$. Expanding the combinations: $\frac{n!}{3!(n-3)!} > \frac{n!}{2!(n-2)!}$. Simplifying the expression: $\frac{1}{3(n-3)!} > \frac{1}{2(n-2)(n-3)!}$. $\frac{1}{3} > \frac{1}{2(n-2)}$. $2(n-2) > 3$. $2n - 4 > 3$. $2n > 7$. $n > 3.5$. Since $n$ must be an integer and $n \geq 3$ for the combinations to be defined,the smallest integer $n$ satisfying the inequality is $n = 4$. However,checking the original expression for $n=4$: ${ }^3 C_2 + { }^3 C_3 = 3 + 1 = 4$ and ${ }^4 C_2 = 6$. $4 > 6$ is false. Checking $n=5$: ${ }^4 C_2 + { }^4 C_3 = 6 + 4 = 10$ and ${ }^5 C_2 = 10$. $10 > 10$ is false. Checking $n=6$: ${ }^5 C_2 + { }^5 C_3 = 10 + 10 = 20$ and ${ }^6 C_2 = 15$. $20 > 15$ is true. Thus,the least value is $n = 6$.

The least value of $n$ for which ${ }^{(n-1)} C_2 + { }^{(n-1)} C_3 > { }^n C_2$ is

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