71i.e., (x+2)! = 182 xP11 * (x-11)!
I think it should be like this.. (Sorry for mine being wrong)
then,
(x+2)! = 182 x! (x-11)! * (x-11)!
=> (x+2) (x+1) = 182
=> x2 +15x-12x-180=0
=> (x+15) (x-12) = 0
So x= 12
if a denotes the number of permutations of (x+2) things taken all at a time,b the number of permutations of x things taken 11 at a time and c the number of permutations of (x-11) things taken all at a time such that a=182bc, find the value of x.
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2 Answers
71