We have to choose 6 digit number by using 1&8 and it must be divisible by 21,
So, equal number of digits 1 and 8 should be used to form a number divisible by 3 and 7. (Or all the digits are either 1 or 8)
In all cases whatever number is formed using six 1′ or six 8′s or three 1′s + three 8′s, that number can be splited into sum of multiples of 1001,1008,8001,8008 (all of them are divisible by 7). So, all the numbers thus formed will be divisible by 21,
Example: 811188=8001×102+1008×10+1008
Number of such numbers =3!3!6!+2=22
Required Probability, p=2622=3211
∴96p=2211×96=33