Q. 285.0( 3 Votes )

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 = 182 bc, find the value of x.

Answer :

Given: a = 182 bc


To find: value of x


a denotes the number of permutations of (x + 2) things taken all at a time


a = P(x+2, x+2)


{Number of arrangements of n things taken all at a time = P(n, n)}


b denotes the number of permutations of x things taken 11 at a time


b = P(x, 11)


{Number of arrangements of n things taken r at a time = P(n,r)}


c denotes the number of permutations of x – 11 things taken all at a time


c = P(x-11, x-11)


{Number of arrangements of n things taken all at a time = P(n,n)}


According to question:


a = 182 bc


P(x+2, x+2) = 182 × P(x, 11) × P(x-11, x-11)





{ 0! = 1}




(x + 2) (x + 1) = 182


x2 + 2x + x + 2 = 182


x2 + 3x + 2 – 182 = 0


x2 + 3x – 180 = 0


x2 – 12x + 15x – 180 = 0


x(x – 12) + 15(x – 12) = 0


(x – 12) (x + 15) = 0


x = -15 or 12


x = 12 {x cannot hold negative value}


Hence, the value of x is 12


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