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  • If <img alt="(156 !) /(151-r) !:(154 !) /(151-r) !=26020: 1" src="https://entrancecorner.oncodecogs.com/gif.latex?%28156%20%21%29%20/%28151-r%29%20%21%3A%28154%20%21%29%20/%28151-r%29%20%21%3D26020

If (156 !) /(151-r) !:(154 !) /(151-r) !=26020: 1 find r_2^P?

Option: 1

32780


Option: 2

66020


Option: 3

46084


Option: 4

23622


Answers (1)

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}.
 

Posted by

Kshitij

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