SOLUTION: Respected Sir, Please help me to solve: If P(2n-1,n):P(2n+1,n-1) = 22:7 ,find n Thank you in aniticipation

Algebra ->  Permutations -> SOLUTION: Respected Sir, Please help me to solve: If P(2n-1,n):P(2n+1,n-1) = 22:7 ,find n Thank you in aniticipation       Log On


   

Question 742102: Respected Sir,
Please help me to solve:

If P(2n-1,n):P(2n+1,n-1) = 22:7 ,find n
Thank you in aniticipation
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
P(2n-1,n):P(2n+1,n-1) = 22:7

We can replace the colons by division symbols ÷

P(2n-1,n)÷P(2n+1,n-1) = 22÷7

P(2n-1,n)÷P(2n+1,n-1) = 22%2F7

Use the formula

P(a,b) = a%21%2F%28a-b%29%21

P(2n-1,n)÷P(2n+1,n-1) = 22%2F7

%282n-1%29%21%2F%28%282n-1%29-n%29%21÷%282n%2B1%29%21%2F%28%282n%2B1%29-%28n-1%29%29%21 = 22%2F7

Remove the inner parentheses in the denominators

%282n-1%29%21%2F%282n-1-n%29%21÷%282n%2B1%29%21%2F%282n%2B1-n%2B1%29%21 = 22%2F7

Simplify by combining like terms:

%282n-1%29%21%2F%28n-1%29%21÷%282n%2B1%29%21%2F%28n%2B2%29%21 = 22%2F7

On the left side, invert the second fraction and change 
division to multiplication:

%282n-1%29%21%2F%28n-1%29%21×%28n%2B2%29%21%2F%282n%2B1%29%21 = 22%2F7

Write (n+2)! as (n+2)(n+1)n(n-1)!

Write (2n+1)! as (2n+1)(2n)(2n-1)!

%282n-1%29%21%2F%28n-1%29%21×%28+%28n%2B2%29%28n%2B1%29%2An%2A%28n-1%29%21+%29%2F%28%282n%2B1%29%282n%29%282n-1%29%21%29 = 22%2F7

Cancel the (2n-1)!'s

cross%28%282n-1%29%21%29%2F%28n-1%29%21× = 22%2F7

Cancel the (n-1)!'s

cross%28%282n-1%29%21%29%2Fcross%28%28n-1%29%21%29× = 22%2F7

Cancel the n's

cross%28%282n-1%29%21%29%2Fcross%28%28n-1%29%21%29× = 22%2F7

All that's left is

%28%28n%2B2%29%28n%2B1%29%29%2F%28%282n%2B1%292%29 = 22%2F7

Cross-multiply:

7(n+2)(n+1) = 22(2n+1)2

7(n²+3n+2) = 44(2n+1)

7n²+21n+14 = 88n+44

Get 0 on the right side:

7n²-67n-30 = 0

Factor the left side as

(n-10)(7n+3) = 0

Use the zero factor property:

n-10 =  0;  7n+3 = 0
   n = 10;    7n = -3
               n = -3%2F7

The original problem contained permutations,
which involves factorials. Only permutations
and factorials involving whole numbers are 
defined, so we can discard the -3%2F7

Solution:  n = 10

Edwin