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

FIND r if:-

5 p_r= 2 ⁶p_r-1

⁵P_r = 2⁶P_r-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.

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