#### If $(156 !) /(151-r) !:(154 !) /(151-r) !=26020: 1$ find $r_2^P$?Option: 1 32780Option: 2 66020Option: 3 46084Option: 4 23622

To solve the equation $\mathrm{156_{r+5}^P: 154_{r+3}^p=26020: 1}$ and find $\mathrm{r_2^P}$ , we will follow a similar process as before.

Let's start by simplifying the equation:

$\mathrm{156_{r+5}^P: 154_{r+3}^P=26020: 1}$

Rewriting the equation using the permutation formula:
$\mathrm{(156 !) /(156-(r+5) !):(154 !) /(154-(r+3) !)=26020: 1}$
Simplifying further:
$\mathrm{(156 !) /(151-r) !:(154 !) /(151-r) !=26020: 1}$

Next, we can simplify the factorials:
$\mathrm{(156 !) /(151-r) !=26020 \times (154 !) /(151-r) !}$
Cancelling out common terms:
$\mathrm{(156 \times 155\times \ldots \times(151-r) \times(150-r) !) /(151-r) !=26020 \times(154 !)}$
Simplifying:
$\mathrm{156 \times 155 \times \ldots \times (151-r)=26020 \times 154 \text { ! }}$

Now, we can solve for $\mathrm{r_2^P}$. The expression $\mathrm{r_2^P}$ represents the number of permutations of 2 objects taken from a set of r objects, which can be calculated as $\mathrm{r !/ (r-2) ! :}$
$\mathrm{r_2^p=r ! /(r-2) !}$

Since we have simplified the equation to the form
$\mathrm{(156 \times 155 \times \ldots \times(151-r) \times(150-r) !)=26020 \times(154 !)}$ , we can set $\mathrm{r=154}$ and calculate $\mathrm{r_2^p :}$
$\mathrm{r_2^P=154 ! /(154-2) !=154 ! / 152 !=154 \times 153}$

Therefore , $\mathrm{r_2^P=154 \times 153= 23622}$.