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
    $43$

Answer

Correct option: B.
$33$
b
$2 \;\times\; 2 \;\times \;2 \;\times \;2\; \times \;2 \;\times \;2\;=\;64$

Divisible by $21$ when divided by $3$ .

Case - $I$ : All $1 \rightarrow$         $(1)$

Case - $II$ : All $8 \rightarrow$        $(1)$

Case - $III$ : $3$ ones 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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