Question

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 ___.

Answer 1

$2 \times 2 \times 2 \times 2 \times 2 \times 2=64$
Divisible by 21 when divided by $3$ .
Case $- I :$ All $1 \rightarrow (1)$
Case - II : All $8 \rightarrow (1) $
Case - III : 3 ones & 3 eights
$\frac{6 !}{3 ! \times 3 !}=20$
Required probability $\therefore p =\frac{22}{64}$
$96 p=96 \times \frac{22}{64}=33$