Dear student
 
For divisibility by 11
 
(Sum of digits at odd places) - (sum of digits at even places ) should be divisible by 11
 
(33+q) - (22+p)  = 11 +(q-p) should be divisible by 11
 
Hence p = q
 
Hence (1,1) (2,2)(3,3).....(9,9) can exist
						

							
To solve this, you need to know the divisibility rules of 8 and 9
 
Divisibility rule for 8
If the sum of the digits of the number is divisible by 8, then the number is divisible by 8                                                              => 7+7+4+9+5+8+P+9+6+Q is divisible by 8 
=> 55 + P + Q is divisible by 8
 
Divisibility rule for 9
For a number to be divisible by 9, the last digit should be divisible by 9
=> Last digit of the number should be 0 or 9
=> Q = 0 or 9
Now, if Q = 0
=> 55 + P should be divisible by 8
=> P = 1 or 9 or 17 or 25
=> There are four possible pairs of {P, Q}
If Q = 9
=> 55 + P + Q should be divisible by 8
=>64+ P should be divisble by 8
=> P = 0 or 8 or 16
=> There are three possible pairs of {P, Q}
Total possible pairs of {P, Q} are 4+3 = 7
Those pairs are {1, 0} {9, 0} {17, 0} {25, 0} {0, 9} {8,9} {16,9}