If {}^{n-1}C_3 + {}^{n-1}C_4 {}^{n}C_3, the | vedclass

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(D) Using the Pascal's identity, ${}^{n-1}C_r + {}^{n-1}C_{r-1} = {}^{n}C_r$, we have: ${}^{n-1}C_3 + {}^{n-1}C_4 = {}^{n}C_4$. Given the inequality:

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Using the Pascal's identity, ${}^{n-1}C_r + {}^{n-1}C_{r-1} = {}^{n}C_r$, we have: ${}^{n-1}C_3 + {}^{n-1}C_4 = {}^{n}C_4$. Given the inequality: ${}^{n}C_4 > {}^{n}C_3$. Expanding the combinations: $\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$. Simplifying the factorials: $\frac{1}{4(n-4)!} > \frac{1}{(n-3)(n-4)!}$. $\frac{1}{4} > \frac{1}{n-3}$. Since $n-3 > 0$, we get: $n-3 > 4$, which implies $n > 7$. Thus, $n$ is just greater than $7$.

If ${}^{n-1}C_3 + {}^{n-1}C_4 > {}^{n}C_3$, then $n$ is just greater than which integer?

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