MCQ
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
- A$30$
- ✓$33$
- C$40$
- D$43$
Divisible by $21$ when divided by $3$ .
Case - $I$ : All $1 \rightarrow$ $(1)$
Case - $II$ : All $8 \rightarrow$ $(1)$
Case - $III$ : $3$ ones and $3$ eights
$\frac{6 !}{3 ! \times 3 !}=20$
Required probability $\therefore p =\frac{22}{64}$
$96 p =96 \times \frac{22}{64}=33$
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