The least value of the natural number n satis | vedclass

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(B) Using the identity $C(n, r) + C(n, r-1) = C(n+1, r)$,we have $C(n, 5) + C(n, 6) = C(n+1, 6)$.Given the inequality $C(n+1, 6) > C(n+1, 5)$.Expandin

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$11$

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(B) Using the identity $C(n, r) + C(n, r-1) = C(n+1, r)$,we have $C(n, 5) + C(n, 6) = C(n+1, 6)$.Given the inequality $C(n+1, 6) > C(n+1, 5)$.Expanding the combinations: $\frac{(n+1)!}{6!(n+1-6)!} > \frac{(n+1)!}{5!(n+1-5)!}$.Simplifying: $\frac{1}{6!(n-5)!} > \frac{1}{5!(n-4)!}$.$\frac{1}{6 	imes 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$.$\frac{1}{6} > \frac{1}{n-4}$.$n-4 > 6$.$n > 10$.Since $n$ is a natural number,the least value of $n$ is $11$.

The least value of the natural number $n$ satisfying $C(n, 5) + C(n, 6) > C(n+1, 5)$ is

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