hi guys!!
want ur help plzzz
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hi ....
here is your answer.
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.
I hope this helps you
here is your answer.
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.
I hope this helps you
khushiagarwal239:
very nice
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