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________.

$n\left(S\right)=2\times 2\times 2\times 2\times 2\times 2=64$

Divisible by 21 when divided by 3 .

Case - I : All 1 $\to$    (1)

Case - II : All 8 $\to$    (1)

Case - III : 3 one & 3 eights

$\frac{6!}{3!\times 3!}=20$

Required probability $\therefore p=\frac{22}{64}$

$96p=96\times \frac{22}{64}=33$