Q100 / 150 · MHT CET 2021 (23 Sep Shift 1) · Mathematics
The difference between the maximum values of ${ }^6 \mathrm{C}_{\mathrm{r}}$ and ${ }^n \mathrm{C}_{\mathrm{r}} 16$, then $\mathrm{n}=$
- A3
- B5
- C2
- Correct answerD4
Solution
The maximum value of ${ }^6 \mathrm{C}_{\mathrm{r}}$ occurs at $\mathrm{r}=\frac{6}{2}=3$
$$
\therefore{ }^6 \mathrm{C}_3=\frac{6 !}{3 ! 3 !}=\frac{6 \times 5 \times 4}{6}=20
$$
As per data given $\left|{ }^6 \mathrm{C}_3-{ }^{\mathrm{n}} \mathrm{C}_3\right|=16$
$$
\therefore{ }^{\mathrm{n}} \mathrm{C}_3=20+16 \text { or }{ }^{\mathrm{n}} \mathrm{C}_3=20-16
$$
If ${ }^n C_3=36 \Rightarrow \frac{n !}{(n-3) ! n !}=36 \Rightarrow n(n-1)(n-2)=216$ is not possible for
$$
\mathrm{n} \in \mathrm{N}
$$
$$
\begin{aligned}
& \text { If }{ }^{\mathrm{n}} \mathrm{C}_3=4 \Rightarrow \frac{\mathrm{n} !}{(\mathrm{n}-3) ! 3 !}=4 \Rightarrow \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)=24 \\
& \Rightarrow \mathrm{n}=4
\end{aligned}
$$
Hence, the correct answer is option: 'D'.