Least value of a natural number n such that left(frac{n-1}{5}right)+left(frac{n-1}{6}right) &l

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6.Permutation and Combination

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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

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