Question

If 'a' denotes the number of permutation of (x+2) things taken all at a time , 'b' the number of permutation of x things taken 11 at a time and 'c' the number of permutation of (x-11) things taken all at a time such that a= 182 bc , find the value of x

Answer 1

Answer:

Answer: (2) 12

Solution: a denotes the number of permutations of x + 2 things taken all at a time, then

a = x+2Px+2 = x+2!

b the number of permutations of x things taken 11 at a time, b = xP11 = x!/(x-11)!

c the number of permutations of x – 11 things taken all at a time, c = x-11Px-11 = (x-11)!

Now, a = 182 bc

Putting the values of a, b and c, we get;

(\mathrm{x}+2) !=182 \times \frac{\mathrm{x} !}{(\mathrm{x}-11) !} \times(\mathrm{x}-11) !(x+2)!=182×

(x−11)!

x!

×(x−11)!

(x+2)!=182x!

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

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

x2+3x+2=182

x2+3x-180=0

(x+15)(x-12)=0

Therefore, x=12,-15