Let X be the set of all five digit numbers formed using 1, 2, 2, 4, 4, 4, 0. For example 22440  is in X while 02244 and 22244 are not in X. Suppose that each element of X has an equal chance of being chosen. Let P be the conditional probability that an element chosen at random is a multiple of 20 given that it is a multiple of 5. Then the value of 76P – 53 is equal to _______

A $\to$event that 5 digit number is divisible by 5 last digit is 0 in all the numbers.
Number of ways = Coefficient of  $x^4$  in  $4!(x^0+x^1)(x^0+\frac{x^1}{1!}+\frac{x^2}{2!})(x^0+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!})=38$
B$\to$ event that 5 digit number is divisible by 20 last 2 digits are 20 (or) 40.
Number of ways = 
Coefficient of  $x^3$  in  $3!(x^0+x^1)(x^0+x^1)(x^0+x^1+\frac{x^2}{2!}+\frac{x^3}{3!})$  +
 Coefficient of  $x^3$  in  $3!(x^0+x^1)(x^0+x^1+\frac{x^2}{2!})(x^0+x^1+\frac{x^2}{2!})=13+18=31$
 $P=P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}=\frac{31}{38}$
 $76P-53=76\left(\frac{31}{38}\right)-53=62-53=9$