Question

The least value of a natural number n such that $(\frac{n-1}{5})+(\frac{n-1}{6})<(\frac{n}{7})$, where $(\frac{n}{r})=\frac{n!}{(n-r)!r!}$, is?

• A. $12$
• B. $13$
• C. $14$
• D. $15$

Answer 1

Answer: (C)

$14$

${\Rightarrow }^{n-1}{C}_5{+}^{n-1}{C}_6<\frac{n}{7}{}^{n-1}{C}_6$

${\Rightarrow }^{n-1}{C}_5<{}^{n-1}{C}_6\left(\frac{n-7}{7}\right)$

$\Rightarrow \frac{(n-1)!}{5!(n-6)!}\cdot \frac{6!(n-7)!}{(n-1)!}<\frac{n-7}{7}$

$\frac{6}{n-6}<\frac{n-7}{7}$

$\Rightarrow 6\times 7<(n-7)(n-6)$

Thus, for the maximum value of $n$,

$(n-7)(n-6)=7\times 8\Rightarrow n=14$