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6.Permutation and Combination
medium
The least value of a natural number $n$ such that $\left(\frac{n-1}{5}\right)+\left(\frac{n-1}{6}\right) < \left(\frac{n}{7}\right)$, where $\left(\frac{n}{r}\right)=\frac{n !}{(n-r) ! r !}, i$
A
$12$
B
$13$
C
$14$
D
$15$
(KVPY-2017)
Solution
(c)
Given,
$\frac{{ }^{n-1} C_5+{ }^{n-1} C_6 < { }^n C_7}{{ }^n C_6 < { }^n C_7}$
$\left[\because{ }^n C_{r-1}+{ }^n C_r={ }^{n+1} C_r\right]$
$\frac{n !}{(n-6) ! 6 !} < \frac{n !}{(n-7) ! 7 !}$
$n-6 > 7$
$n>13$
$\therefore$ Least value of $x=14$
Standard 11
Mathematics
