Question

If $2nC_{3}:^{n}C_{2} = 44:3$, then for which of the following values of $r$, the value of $nC_{r}$ will be 15.

• A. $r = 3$
• B. $r = 4$
• C. $r = 6$
• D. $r = 5$

Answer 1

Answer: (B)

$r = 4$

Answer 2

$\frac{(2n)\mspace{6mu}!}{(2n - 3)\mspace{6mu}!\mspace{6mu}.\mspace{6mu} 3\mspace{6mu}!} \times \frac{2\mspace{6mu}!\mspace{6mu} \times (n - 2)\mspace{6mu}!}{n\mspace{6mu}!}\mspace{6mu} = \frac{44}{3}$

$\Rightarrow$ $n$

$\Rightarrow 4(2n - 1) = 44 \Rightarrow 2n = 12 \Rightarrow n = 6$

Now $6C_{r} = 15 \Rightarrow 6C_{r} =^{6} ⥂ C_{2}$ or $6C_{4} \Rightarrow r = 2,\mspace{6mu} 4$.