The least value of natural number n satisfyin | vedclass

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(A) Using the Pascal's identity,$^nC_r + ^nC_{r-1} = ^{n+1}C_r$,we have $^nC_6 + ^nC_5 = ^{n+1}C_6$.Given the inequality: $^{n+1}C_6 > ^{n+1}C_5$.Expa

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) Using the Pascal's identity,$^nC_r + ^nC_{r-1} = ^{n+1}C_r$,we have $^nC_6 + ^nC_5 = ^{n+1}C_6$.Given the 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 the factorials:$\frac{1}{6 	imes 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$.$\frac{1}{6} > \frac{1}{n-4}$.Since $n$ is a natural number,$n-4 > 6$,which implies $n > 10$.The least natural number $n$ satisfying this condition 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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