The least value of natural number n satisfyin | vedclass

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(A) Using the Pascal's identity,$C(n, r) + C(n, r-1) = C(n+1, r)$,we have:$C(n, 6) + C(n, 5) = C(n+1, 6)$Given inequality: $C(n, 5) + C(n, 6) > C(n+1,

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$11$

Vedclass · Answer

(A) Using the Pascal's identity,$C(n, r) + C(n, r-1) = C(n+1, r)$,we have:$C(n, 6) + C(n, 5) = C(n+1, 6)$Given inequality: $C(n, 5) + C(n, 6) > C(n+1, 5)$Substituting the identity: $C(n+1, 6) > C(n+1, 5)$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)!}$$\frac{1}{6 	imes 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$$\frac{1}{6} > \frac{1}{n-4}$$n - 4 > 6$$n > 10$The least natural number $n$ satisfying this is $11$.

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

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