MCQ
A five digit number divisible by $3$ has to formed using the numerals $0, 1, 2, 3, 4$ and $5$ without repetition. The total number of ways in which this can be done is
  • $216$
  • B
    $240$
  • C
    $600$
  • D
    $3125$

Answer

Correct option: A.
$216$
a
(a) We know that a five digit number is divisible by $3$, if and only if sum of its digits $(= 15)$ is divisible by $3$, therefore we should not use $0$ or $3$ while forming the five digit numbers.

Now, $(i)$ In case we do not use $0$ the five digit number can be formed (from the digit $1, 2, 3, 4, 5)$ in $^5{P_5}$ ways.

$(ii)$ In case we do not use 3, the five digit number can be formed (from the digit $0, 1, 2, 4, 5$) in ${^5}{P_5}{ - ^4}{P_4}=5!-4!=120-24-96$ ways.

$\therefore $ The total number of such $5$ digit number

$ = {\,^5}{P_5} + {(^5}{P_5}{ - ^4}{P_4}) = 120 + 96 = 216$.

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