The least value of the natural number n satis | vedclass

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(B) Using the Pascal's identity,${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$,we have ${}^nC_5 + {}^nC_6 = {}^{n+1}C_6$. Given inequality: ${}^{n+1}C_6 > {}^{

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) Using the Pascal's identity,${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$,we have ${}^nC_5 + {}^nC_6 = {}^{n+1}C_6$. Given inequality: ${}^{n+1}C_6 > {}^{n+1}C_5$. Expanding the combinations: $\frac{(n+1)!}{6!(n-5)!} > \frac{(n+1)!}{5!(n-4)!}$. Dividing both sides by $(n+1)!$ and simplifying: $\frac{1}{6!(n-5)!} > \frac{1}{5!(n-4)!}$. Since $6! = 6 	imes 5!$ and $(n-4)! = (n-4) 	imes (n-5)!$,we get: $\frac{1}{6 	imes 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$. Canceling common terms: $\frac{1}{6} > \frac{1}{n-4}$. This implies $n-4 > 6$,so $n > 10$. The least natural number $n$ satisfying this is $n = 11$.

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

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