Q57 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics

The least value of $n$ such that ${ }^{(n-1)} C_6+{ }^{(n-1)} C_7 < { }^n C_8$ is

  • A
    14
  • B
    15
  • Correct answer
    C
    16
  • D
    117

Solution

${ }^{(n-1)} C_6+{ }^{(n-1)} C_7 < { }^n C_8$ $\Rightarrow{ }^n C_7 < { }^n C_8$ $\Rightarrow{ }^n C_7-{ }^n C_8 < 0$ $\Rightarrow \quad \frac{n !}{7 !(n-7) !}-\frac{n !}{8 !(n-8) !} < 0$ $\Rightarrow \frac{n !}{7 !(n-8) !}\left[\frac{1}{n-7}-\frac{1}{8}\right] < 0$ $\Rightarrow \frac{8-(n-7)}{8(n-7)} < 0$ $\Rightarrow 15-n < 0 \Rightarrow n>15$ $\therefore \quad$ The least $n$ satisfying above condition is $n=16$.

Hence, the correct answer is option: 'C'.

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