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.
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