Question

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, then the value of x is

• A. 15
• B. 12
• C. 10
• D. 18

Answer 1

Answer: (B)

12

Answer 2

We have a = ^{x + 2}P_{x + 2} = (x + 2)!, b = ^xP₁₁ = $\frac{x!}{(x - 11)!}$ and

c = ^{x – 11}P_{x – 11} = (x – 11)!

Now a = 182 bc ⇒ (x + 2)! = 182. $\frac{x!}{(x - 11)!}$ (x – 11)!

⇒ (x + 2)! = 182 x! ⇒ (x + 2) (x + 1) = 182 ⇒ x = 12.