How many 6 digits number will be possible which cannot be divided by 9 and 5 its first digit and 4 is the third digit

Soln plzz

Ye solve kro..

Case 1)

(x - 2) = (y - 3) = (z - 4) = - 1

x = 1, y = 2, z = 3

Case 2)

(x - 2) = - 1, (y - 3) = (z - 3) = 1

x = 1, y = 4, z = 5

Case 3)

(x - 2) = (z - 4) = 1, (y - 3) = -1

x = 3, y = 2, z = 5

Case 4)

(x - 1) = (y - 3) = 1, (z - 4) = -1

x = 3, y = 4, z = 3

p = 8, q = 12, r = 16

p + q - r = 4

At n=2. Expression 3 (x^2+3x+2)(x+3)
2x+9x=11x=11

6*(6+3+2)/6=11. At n=3
Option d

Sir how to solve such type questions ?

(1, 2), (1, 4), (1, 6), (2, 3), (2, 5), (2, 9), (3, 4), (3, 8), (4, 7), (4, 9), (5, 6), (5, 8), (6, 7), (8, 9)


14/36

7/18