Question

if 5pr = 2, 6pr-1 find r

Answer 1

5Pr = 5!/(5-r)!, and 2(6P(r-1)) = 2*6!/(6-r+1)! so 

5!(6-r+1)! = 2*6!(5-r)! 
6*5!(7-r)! = 6*2*6!(5-r)! 
6!(7-r)! = 12*6!(5-r)! 
12 = 6!(7-r)! / 6!(5-r)! 
12 = (7-r)(6-r) 

so the solutions are r = 3 and r = 10, though the latter gives us 5P10 = 2*(6P9) which is 0=0, and therefore redundant as 0≤r≤n for Permutations.

Answer 2

Answer:10 or 3

Step-by-step explanation:

5Pr=2 6Pr-1

5!/(5-r)! = 2.6!/(6-(r-1))!

We have 6! = 5!.6

1/(5-r)! = 2*6/(6-r+1)!

1/(5-r)! = 2*6/(7-r)!

We know,

(7-r)!=(5-r)! (6-r)* (7-r)

i.e 1= 12/(6-r)(7-r)

6-r(7-r)=12

From this equation we will get,

r^2+13r-30=0

By solving this quadratic equation we will get,

r=3 or r=10