sum of a five digit number is 41 find the probability such that the number is divisible by 11

In order to get the sum as 41, the following 5 digit combination exist: 
99995--->number of 5 digits =5 
99986--->number of 5 digit =20 
99977--->number of 5 digit =10 
99887--->number of 5 digit =30 
98888--->number of 5 digit =5 
Now, 70 such number exists. 
Now for a 5 digit number of form (pqrst) to be divisible by 
11(p+r+t)-(q+s)=11, also (p+r+t)+(q+s)=41 
p+r+t=26, q+s=15 
(p,r,t)=(9,9,8) and (q,s)=(8,7) --------(i) 
or (p,r,t)=(9,9,8) and (q,s)=(9,6) --------(ii) 
Using 1st equation we can construct 3!/2! ×2!=6 numbers. 
Using 2nd equation we can construct 3!/2! ×2!=6 numbers. 
Number of favorable cases = 12. 
Hence, required probability = 12/70 = 6/35