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Q.
Five digit number divisible by $3$ is formed using $0, 1, 3, 4, 6, 7$ without repetition. Total number of such numbers are
Permutations and Combinations
Solution:
A number is divisible by $3$ if sum of its digits be divisible by $3$.
Hence, the selection of $5$ digits for forming five digit numbers divisible by $3$
are either $ 1, 3,4,6, 7$ or $0,1,4,6, 7$ or $0,1,3,4,7$. In first case i.e. when $1, 3, 4, 6, 7$ are taken as digits, the numbers of five digit numbers will be $ 5! = 120$. In rest of the two cases, we have $4 \times 4! = 96$ in each case.
$\therefore $ Total number of required numbers
$ = 120 + (2 \times 96) = 312$