FIND r if:-

5 pr= 2 6pr-1

5Pr = 2 6Pr-1

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

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

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

== 1 = 12 / (7-r) (6-r) [See I have cancelled out terms from L.H.S. AND R.H.S.]

== (7-r) (6-r) = 12

==42 - 7r -6 r + r2 = 12

==r2 - 13r + 42 -12 =0

== r2 - 13r +30 =0

==r2 -10r -3r + 30 = 0

== r( r-10) - 3( r-10)=0

== (r-10) (r-3) =0

== (r-10) = 0 or r-3 =0

== r = 10 or r = 3

So, r = 3 only because the value of r cant be higher than 5 and 6 as they are the no. out of which we have to take arrangement, so 10 is neglected.

  • 13

5Pr = 2 6Pr-1

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

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

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

== 1 = 12 / (7r) (6r) [See I have cancelled out terms from L.H.S. AND R.H.S.]

== (7r) (6r) = 12

==42 7r 6 r + r2 = 12

==r2 13r + 42 12 =0

== r2 13r +30 =0

==r2 10r 3r + 30 = 0

== r( r10) 3( r10)=0

== (r-10) (r-3) =0

== (r-10) = 0 or r-3 =0

== r = 10 or r = 3

So, r = 3 only because the value of r cant be higher than 5 and 6 as they are the no. out of which we have to take arrangement, so 10 is neglected.

  • 2

how this 3rd step came i didn't understand

  • 2

see new one..or second post...in first post there is no negative sign..

  • 1

understood or not???

  • 2

understood thanks for taking so much efforts

  • 1

6th step?

  • 1

right the step please....

  • 0

write the step please...

  • 0

42 7r 6 r + r2 = 12

  • 0

I said na... see my second post...see that only and don't be confuse...i have just simply multiplied....both (7-r) and (6-r)

  • 4

ok ok

  • 2
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