If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is $p$, then $96 p$ is equal to ......

Total possibilities to make a 6-digit number using only 1 and 8
Every position has 2 possibility either 1 or 8 $\therefore 2^6$ Number divisible by 21 : 1. for divisibility by 3
A (sum of digits must be divisible by 3 ) (a) Number of 1 's $=$ Number of 8 (b) Every digit is 1 (c) Every digit is 8. 2. for divisiblity by 7. |2 (last digit) - remaining númber $\mid=7 \mathrm{k}$ $k \in Z$ (a) all are $1 \mathrm{~s} \Rightarrow 1$ no (b) all are $8 s \Rightarrow 1$ no (c) 31 's and $38 \mathrm{~s} \Rightarrow \frac{6!}{3!3!}$
Total number divisible by 21 = number divisible by 3 and 7 $\begin{aligned} & =\frac{6!}{3!3!}+2=\left(\frac{6 \times 5 \times 4 \times 3!}{3!\times 3 \times 2}\right)+2=22 \\ \therefore \text { Probability } & =\frac{22}{2^6}=p \\ 96 p & =96 \times \frac{22}{2^6}=96 \times \frac{22}{2 \times 2 \times 2 \times 2 \times 2 \times 2}=33 \end{aligned}$