Prove the following

Question:

If ${ }^{n} P_{r}={ }^{n} P_{r+1}$ and ${ }^{n} C_{r}={ }^{n} C_{r-1}$, then the value of $r$ is equal to:

  1. 1

  2. 4

  3. 2

  4. 3


Correct Option: , 3

Solution:

${ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}+1} \Rightarrow \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}-1) !}$

$\Rightarrow(\mathrm{n}-\mathrm{r})=1$            ..........(1)

${ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{r-1}$

$\Rightarrow \frac{n !}{r !(n-r) !}=\frac{n !}{(r-1) !(n-r+1) !}$

$\Rightarrow \frac{1}{r(n-r) !}=\frac{1}{(n-r+1)(n-r) !}$

$\Rightarrow \mathrm{n}-\mathrm{r}+1=\mathrm{r}$

$\Rightarrow \mathrm{n}+1=2 \mathrm{r}$  .............(2)

$(1) \Rightarrow 2 r-1-r=1 \Rightarrow r=2$

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